I need to prove the exponent identity $(a^b)^c = a^{bc}$, where $a,b,c \in \mathbb{Z}$.
Apparently this proof is elementary/trivial, but I can't think of how to prove it. I need it as a lemma for another problem. Can anyone please point me in the right direction or show me why this is true?
On
Let's prove it, first, for $c\ge0$. Then the assertion is obvious for $c=0$. Assume it holds for $c\ge0$: $$ (a^b)^{c+1}=(a^b)^c\cdot a^b=a^{bc}\cdot a^b=a^{bc+b}=a^{b(c+1)} $$ so, by induction the assertion is proved for $c\ge0$. If $c<0$, then, setting $c=-d$, we have $$ (a^b)^c=\frac{1}{(a^b)^d}=\frac{1}{a^{bd}}=\frac{1}{a^{-bc}}=a^{bc} $$
Fill in the details.
Presumably by definition you know that $a^n \cdot a = a^{n+1}$.
Use this to prove that $a^n \cdot a^m = a^{n+m}$.
Then you have $$(a^n)^k = \underbrace{a^n \cdot a^n \cdot \cdots \cdot a^n}_{k\ \text{times}} = a^{\overbrace{n+n+\cdots+n}^{k\ \text{times}}} = a^{nk}$$
There are some details to be filled in, e.g. what happens when $k$ is negative, but this should help you get the idea.