Prove $a^{n}*a^{m}=a^{n+m}$ $a^{n}*a^{m}=a^{n+m}$

Induccion over $a$

Being $a=0$ $$0^{n}*0^{m}=0^{n+m}$$
            $$0*0=0$$ 
it works!

Hip. $$a^{n}*a^{m}=a^{n+m}$$

Now let $S(n)$ be the succesor of $n$
We have to show the following: $$S(a)^{n}*S(a)^{m}=S(a)^{n+m}$$

I been trying to develop the left side of the equality, but seems a little obious to me the conection with the right side. Does any suggestion of whats could go in the middle?
Induct on $m$ instead. The case $m=0$ is trivial by $a^0=1$. If the case $m=k$ works, $a^na^{k+1}=a^naa^k=a^{n+1}a^k=a^{n+1+k}=a^{n+k+1}$.
Induction over $a$ is terrible, if doable at all.

Easy over $m$:

$$a^na^1=a^{n+1}$$ by definition of the integer powers, and

$$a^{n}a^{m+1}=a^n(a^ma)=(a^na^m)a=a^{n+m}a=a^{n+m+1}=a^{n+(m+1)}.$$