There is a rule in the laws of exponents that students routinely confuse with the product law: (a^m)^n = a^{m \cdot n}. The exponents multiply. Why multiply and not add? Because the expression is telling you something very specific — it is telling you to take the block a^m and repeat that block n times. Each block is itself m copies of a. So the question "how many a's are there in total?" is the same as asking "how many items are in n boxes of m items each?" You already know the answer to that from primary-school multiplication. It is m \cdot n.
Everything on this page is a single picture of that fact. You will see n boxes on a canvas, each containing m tiles marked a. When you press Unzip, each box opens and its tiles spill out into a single long row. Count the row. You will always find m \cdot n tiles. The formula (a^m)^n = a^{m \cdot n} is not a law you memorise; it is the number of tiles on the tray after the zipper has opened.
The widget
Two sliders set the inner and outer exponents. The inner exponent m is how many a's live inside each box. The outer exponent n is how many boxes there are. When the animation starts, each box widens and its boundary dissolves, leaving its tiles to merge into the single long row beside the other boxes' tiles. The expression at the top of the canvas switches from (a^m)^n to a^{m \cdot n} the moment the zipper finishes. The readout below the canvas always shows the algebraic identity.
The two colours are a visual aid. Tiles from even-numbered boxes (the first, the third, the fifth) are blue; tiles from odd-numbered boxes are orange. This way, after unzipping, you can still see which tiles came from which original block, which is useful when you want to convince yourself that nothing was added or lost in the process.
Try these
m = 3, n = 2, the default. Two boxes, each with three a's. Unzip and you see six a's in a row, alternating blue and orange in a block-of-three pattern. The readout: (a^3)^2 = a^6.
m = 2, n = 3. Three boxes of two a's. Unzip, and you again get six a's — but now in a blue-orange-blue pattern of pairs. The total is the same as before, because 2 \cdot 3 = 3 \cdot 2 = 6. This is the visual proof that the order of exponents in a power-of-a-power does not matter: (a^m)^n = (a^n)^m.
m = 4, n = 1. One box containing four a's. Unzip and… almost nothing happens, because there is only one box, and its contents were already a row of four. The readout: (a^4)^1 = a^4. An outer exponent of 1 means "take one copy of what's inside" — and one copy of anything is just itself.
m = 1, n = 4. Four boxes, each holding a single a. Unzip and the four singletons line up as a \cdot a \cdot a \cdot a. The readout: (a^1)^4 = a^4. An inner exponent of 1 means each box is a lonely a, and then stacking four of those together is simple multiplication.
m = 4, n = 4. Four boxes of four a's each. This is the largest the widget can show — unzip it and you get sixteen a's, neatly lined up in alternating colour groups of four. The readout: (a^4)^4 = a^{16}. Beyond this, the tiles would not fit on a phone screen, which is why the widget caps out at 4 \times 4.
What you are actually seeing
Read the algebra exactly as it is written and the picture writes itself. The outer exponent n on the expression (a^m)^n means "take the thing in brackets and multiply it by itself n times." The thing in brackets is a^m, which is itself a shorthand for a \cdot a \cdot \dots \cdot a with m factors. So the full expansion, before any simplification, is n copies of "a \cdot a \cdot \dots \cdot a" — and each of those copies has m factors of a inside it.
Now you just count. Total number of a's multiplied together = (number of boxes) × (factors per box) = n \times m = m \cdot n. And m \cdot n factors of a multiplied together is, by the definition of exponentiation, a^{m \cdot n}. This is the whole proof. The widget is a mechanical enactment of that counting argument — it takes the (a^m)^n arrangement, physically removes the box boundaries one by one, and leaves you with a single row of m \cdot n factors to count.
A sanity check
Take (a^2)^2 and write out every factor by hand. The inside is a^2 = a \cdot a. The outside is "do that twice," so (a^2)^2 = (a \cdot a) \cdot (a \cdot a) = a \cdot a \cdot a \cdot a = a^4. Count them: four. And 2 \cdot 2 = 4. The formula matches what you just counted.
Do it again for (a^3)^2. Inside: a \cdot a \cdot a. Outside: twice. So (a^3)^2 = (a \cdot a \cdot a) \cdot (a \cdot a \cdot a) = a^6. Count: six. And 3 \cdot 2 = 6. Matches. Every example you can think of reduces to this counting exercise, and the widget does it for you.
Common confusion — (a^m)^n vs a^{m^n}
Without brackets, the expression a^{m^n} does not mean the same thing as (a^m)^n. Mathematicians agree on a convention for expressions with stacked exponents: read them top-down, right to left — in other words, the topmost exponent applies first. This is called right-associative evaluation. So a^{m^n} means a^{(m^n)}, not (a^m)^n.
The numerical gap between the two can be enormous. Take a = 2, m = 3, n = 2. The expression 2^{3^2} means 2^{(3^2)} = 2^9 = 512. The expression (2^3)^2 means 8^2 = 64. Same three numbers, same operations, different groupings — and the first is eight times the second. In expressions with stacked exponents, the brackets are not optional decoration; they change the answer by orders of magnitude.
So whenever you see a power-of-a-power in an algebra problem, check whether the bracket sits around the inner power or around the whole thing. The widget on this page only visualises the (a^m)^n case, because the a^{m^n} case is a genuinely different operation and would need a different picture.
Extends beyond integers
The rule (a^m)^n = a^{m \cdot n} does not stop at integer exponents. It remains true when m and n are rational numbers, and when they are real numbers more generally — for example, (a^{1/2})^2 = a^{1 \cdot (1/2) \cdot 2} = a^1 = a, which is the algebraic way of saying "the square of the square root of a is a." The widget's "count the tiles" story breaks down once you leave the integers, because you cannot have half a tile in a box. But the algebraic identity carries through, and its proof in that general setting uses properties of the exponential and logarithm functions rather than a counting argument.
For the companion picture of the product rule a^m \cdot a^n = a^{m+n} — where stacks of tiles merge instead of boxes unzipping — see animated product rule: stacks merge. The two widgets together cover the two most-confused laws: multiply when you have a power of a power, add when you have a product of powers with the same base.
Closing
The zipper is not a proof in the formal sense. It is a picture. But it is the picture that makes the proof obvious. The outer exponent multiplies because it is telling you to repeat the inner group — and when you repeat a group of m things a total of n times, you get m \cdot n things. Anything else would be wrong. You do not need to memorise (a^m)^n = a^{m \cdot n}; you need to see that the right-hand side is just what the left-hand side counts.