Here is a tiny animation that will put the question "why is a^0 = 1?" to bed permanently. Look at the tower of powers of 2, walk down it one step at a time, and watch what happens on the way to 2^0.

Going up the ladder, each step doubles: 2^1 = 2, 2^2 = 4, 2^3 = 8, each one twice the last. Reverse the direction, and each step must halve — because what doubling does, halving undoes. So starting at 2^3 = 8 and stepping down:

2^3 = 8 \;\longrightarrow\; 2^2 = 4 \;\longrightarrow\; 2^1 = 2 \;\longrightarrow\; 2^0 = \;?

Each arrow halves. Follow the halving one more step and the answer writes itself: 2^0 = 1.

The interactive staircase

Drag the slider below from n = 4 down to n = -2. Each step halves the previous value. The number at the bottom of the chain is the value of 2^n at whatever n you stopped at. When you land on n = 0, the value is 1 — not by decree, but because the halving pattern leaves no other option.

Halving staircase from two cubed down to two to the minus two A draggable point on a curve of two to the n, with n ranging from negative two to four. At n equals three, the value is eight. At n equals two, four. At n equals one, two. At n equals zero, one. At n equals negative one, one half. At n equals negative two, one fourth. Each step to the left halves the previous value. A readout at the top shows the current n and the current two to the n. -2 0 1 2 3 4 (n) 0 1 12 16 (2ⁿ) ↔ drag to step down through zero
The curve $y = 2^n$ from $n = -2$ to $n = 4$. Each integer step to the right *doubles* the value; each step to the left *halves* it. The curve passes smoothly through $n = 0$, where the value is exactly $1$. To the left of $n = 0$, the halving continues: $2^{-1} = \tfrac{1}{2}$, $2^{-2} = \tfrac{1}{4}$. There is no jump and no special case at $n = 0$.

The quotient-law proof, once more

The pattern above is not a coincidence. It is forced by the quotient law of exponents. Apply the law to a single example: \dfrac{2^3}{2^3}.

\frac{2^3}{2^3} = 2^{3 - 3} = 2^0

Why the quotient law gives 3 - 3: the law says \dfrac{a^m}{a^n} = a^{m-n}, and we are applying it with m = n = 3.

But the left side, \dfrac{2^3}{2^3}, is just \dfrac{8}{8} = 1, because any nonzero number divided by itself is 1. So the two sides of the equation force 2^0 = 1. This is not a convention. It is the only value of 2^0 that keeps the quotient law consistent, and the quotient law is too useful to give up.

The same argument works for any nonzero base: \dfrac{a^n}{a^n} = 1 and \dfrac{a^n}{a^n} = a^{n-n} = a^0, so a^0 = 1 for every a \neq 0. The base does not matter. Seven to the zero is one. A hundred to the zero is one. A million to the zero is one. \pi^0 = 1.

What about 0^0?

You may ask: does the argument break when a = 0? It does. The quotient law requires dividing by a^n, and if a = 0 then a^n = 0 and you are dividing by zero. So the argument says nothing about 0^0.

In school, 0^0 is usually left undefined or sometimes defined as 1 by convention (this choice makes polynomial formulas like \sum a_k x^k evaluate correctly at x = 0, among other things). It is one of those edge cases where different contexts pick different conventions. For every nonzero a, though, a^0 = 1 is forced and there is nothing to debate.

The halving pattern in a table

If the slider feels slippery, here is the same staircase laid out row by row. Each row is obtained from the one above it by halving.

n 2^n
4 16
3 8
2 4
1 2
0 1
-1 \tfrac{1}{2}
-2 \tfrac{1}{4}
-3 \tfrac{1}{8}

Cover the right-hand column and try to fill it in yourself. The rule "halve as you go down" leaves only one possible value for each row — and the n = 0 row is 1, sandwiched between 2 above it and \tfrac{1}{2} below it. The number 1 is the only value that can sit between those two while preserving the halving pattern.

Why this is the best way to remember the fact

Students who memorise "a^0 = 1" as a rule often forget whether it applies to all a, whether it is "because anything to the zero is one" (not a reason), and why it should be true. The halving staircase replaces memorisation with a picture. You do not remember the fact; you re-derive it in two seconds by walking down from 2^3 to 2^0.

It also makes the negative-exponent rule a^{-n} = \tfrac{1}{a^n} obvious: negative exponents are just more halving, continuing the same pattern below zero. They are not a new definition. They are the staircase, extended.

Related: Exponents and Powers · A Tower of 2s: Watch Doubling Explode Off the Screen by Step 30 · Is 2⁻³ Really Negative Eight? The Sign-on-the-Answer Trap · Tile-View Proof of the Three Core Exponent Laws