This is my proof, using multiple induction.
Base Case: $k = l = 1$
$$(a^1)^1 := a^1 := a$$
Assume the inductive hypothesis $a^k a^l = a^{k+l}$
For $k+1$
$$(a^{k+1})^l = (a^k a)^l = (a^k)^l a^l = a^{kl} a^l = a^{kl+l} = a^{(k+1)l}$$
Similarly for $l+1$
$$(a^k)^{l+1} = (a^k)^l (a^k)^1 = (a^{kl}) a^k = a^{kl+k} = a^{k(l+1)}$$
I have already proved addition of powers before this exercise so that is assumed. My only problem is, can I assume $(ab)^l = a^l b^l$ for $a,b$ elements of a group and $l \in \mathbb{Z}$?
Thank you!
I will use this definition: for $n\in\mathbb{Z}^{+}$, $x^{n}=x\cdots x$ ($n$ terms), $x^{0}=1$, $x^{-n}=x^{-1}\cdots x^{-1}$ ($n$ terms). I will use this property: $(ab)^{-1}=b^{-1}a^{-1}$ for $a,b\in G$. We need a lemma first.
Proof of lemma. We have $$ \begin{split} x^{a}x^{-a}&=(x\cdots x)(x^{-1}\cdots x^{-1}) \quad \textrm{($a$ terms each)}\\ &=1. \end{split} $$ Similarly, $x^{-a}x^{a}=1$. So $(x^{a})^{-1}=x^{-a}$.
Proof of the question.
Case 1: $k,l>0$.
Then $$ \begin{split} a^{k}a^{l}&=(a\cdots a)(a\cdots a) \quad \textrm{($k,l$ terms, respectively)}\\ &=a\cdots a \quad \textrm{($k+l$ terms)}\\ &=a^{k+l}. \end{split} $$
Case 2: $k$ or $l=0$. Easy.
Case 3: $k<0$ and $l>0$.
If $-k> l$, then $$ \begin{split} a^{k}a^{l}&=a^{-(-k)}a^{l}=(a^{-k})^{-1}a^{l} \quad \textrm{by the lemma}\\ &=(a^{l}a^{-k-l})^{-1}a^{l} \quad \textrm{by case 1}\\ &=(a^{-k-l})^{-1}(a^{l})^{-1}a^{l}=a^{-(-k-l)} \quad \textrm{by the lemma}\\ &=a^{k+l}. \end{split} $$ If $-k\leq l$, then $$ \begin{split} a^{k}a^{l}&=a^{-(-k)}a^{l}=(a^{-k})^{-1}a^{l} \quad \textrm{by the lemma}\\ &=(a^{-k})^{-1}a^{-k}a^{l-(-k)} \quad \textrm{by case 1 or case 2}\\ &=a^{k+l}. \end{split} $$
Case 4: $k>0$ and $l<0$. Similar to case 3.
Case 5: $k,l<0$.
Then $$ \begin{split} a^{k}a^{l}&=a^{-(-k)}a^{-(-l)}=(a^{-k})^{-1}(a^{-l})^{-1} \quad \textrm{by the lemma}\\ &=(a^{-l}a^{-k})^{-1}=(a^{-l-k})^{-1} \quad \textrm{by case 1}\\ &=a^{-(-l-k)} \quad \textrm{by the lemma}\\ &=a^{k+l}. \end{split} $$