Can anyone please verify this proof:
Prove that $(g^m)^n = (g)^{mn}$
Proof: $(g^m)^n = \underbrace{g^m*g^m....*g^m}_{\text{n times}}$
$= \underbrace{[\underbrace{(g*g*g...*g)}_{\text{m times}}*\underbrace{(g*g*g...*g)}_{\text{m times}}....\underbrace{(g*g*g...*g)]}_{\text{m times}}}_{\text{n times}}$
Then this is $m$ $g's$ $n$ times
$= g^{mn}$
Does this proof work? If not, then what alternative proof is there? Thanks.
This is the basic idea, yes. You might want to write it out as an induction proof, as an exercise. That's the usual way of taking "$\cdots$" out of proofs.
In this case, it's clear that $(g^{m})^1=g^{m\cdot 1}$ (because both expressions are simply equal to $g^m$). So you can assume that, for some $n$, we have $(g^m)^n=g^{mn}$, and then use that to show that $(g^m)^{n+1}=g^{m(n+1)}$. You'll want to use a theorem telling you that $g^ag^b=g^{a+b}$. Have you got that available?