Proving Power Rules 
Let $M$ be a monoid and $a\in M$, we define in induction $a^{n+1}=a^n\cdot a $ where $a^{1}=a$
Prove: 1. $\forall m,n\in \mathbb{N}$ $a^na^m=a^{n+m}$


*$\forall m,n\in \mathbb{N}$ $(a^n)^m=a^{nm}$


*Assume $a$ has an inverse $\forall m,n\in\mathbb{N}$ $(a^{-1})^n=(a^n)^{-1}$

1.$$a^na^m=\underbrace{a\cdot a \dotsm a}_{n}\underbrace{a\cdot a \dotsm a}_{m}=(\underbrace{a\cdot a \dotsm a}_{n})(\underbrace{a\cdot a \dotsm a}_{m})=a^{m+n}$$
2.$$(a^n)^m=(\underbrace{a\cdot a \dotsm a}_{n})^m=\underbrace{a^m\cdot a^m\dotsm a^m}_{n}={a^{\underbrace{m+m+\dotso +m}_{n}}}=a^{mn}$$


*$$e=a\cdot a\dotsm a \cdot a^{-1}\cdot a^{-1} \dotsm a^{-1}=a^n\cdot (a^{-1})^n$$ so $(a^{-1})^n$ is the inverse of $a^n$ or $(a^{n})^{-1}$
Is those proofs valid?
 A: No. These are informal proofs. Instead, since powers are defined recursively, proofs will be done by induction.
Consider the first rule. We proceed by induction on $m$.
If $m = 1$, then $a^{n+1} = a^{n} \cdot a = a^{n} \cdot a^{1}$ is just a matter of applying the definitions.
If $m > 1$ you have
$$
a^{n + m} = a^{n + (m-1) + 1} = a^{n + (m-1)} \cdot a 
=
a^{n} \cdot a^{m-1} \cdot a
=
a^{n} \cdot a^{m},
$$
where the second and fourth equalities depends on the definitions, and the third one follows from the inductive hypothesis.
A: First of course that depends on what level you want to prove it, but if you accept to resort to hand-wavery then...
...they look a bit backward. 
For the first it more look like the third expression would be a rewrite of the first, the second a rewrite of the third and the last is a rewrite of the second.
For the second it's more forward, but the second equality seem to be missing a step (where you have $m$ chunks of $n$ $a$s each.
The last is the only one that seems alright to me.
If you want to straighten it you could do that quite easily using induction (which is more natural since the definition is recursive). You don't have to realize that $a^n$ is the product of $n$ $a$s then.
