We know that

1. a^n = a^(n-1) * a ==> a^(n + 1) = a^n * a

2. a^0 = 1

as givens.

So n = 0 case:

a^m * a^0 = a^m * 1, by 2

= a^m = a^(m + 0)

Now assume that a^m * a^n = a^(m + n) for some n. We need to show that a^m * a^(n + 1) = a^(m + n + 1).

So

a^m * a^(n + 1) = a^m * a^n * a, by 1

= a^(m + n) * a, by hypothesis

Now, m + n is just some integer x.

= a^x * a = a^(x + 1), by 1

= a^(m + n + 1)

Thus a^m * a^n = a^(m + n) for n = 0 and thus for all integer n >= 0.

-Dan

Frankly proof of this by induction is a bit of overkill as far as I'm concerned. We can prove it to be true just by applying the definition of a^n.