Proving exponent properties in a group I am trying to prove this following statement about groups.

Let $G$ be a group, and $a, b \in \mathbb{Z}$. Let $x \in G$.
(a) Show $(x^a)^{-1} = x^{-a}$.
(b) Show $x^{a+b} = x^a x^b$ and $(x^a)^b = x^{ab}$.
(c) Show that $|x| = |x^{-1}|$.

I am having difficulty proving these sequentially. The proof of (a) I would write, for example, would be
$$x^a x^{-a} = x^{a + (-a)} = x^0 = e,$$
where $e$ is the identity in $G$. Similarly,
$$x^{-a} x^a = x^{-a + a} = x^0 = e.$$
Hence, by uniqueness of the inverse in a group,
$$(x^a)^{-1} = x^{-a}.$$
The problem is that I have used part (b) to prove part (a), and this is almost surely not the way I was supposed to do it. Without using (b), I cannot think of a way to prove (a). A hint would be very much appreciated.
I similarly cannot figure out how to prove (b), as this is a fact I have always assumed to hold and used as an auxilary lemma in proofs of this kind.
I believe I can prove see. I assume that writing $|x|$ implies that the order is finite. Otherwise, if $x$ has infinite order, then certainly $x^{-1}$ has infinite order, because $x^n \neq e$ for any $n$, so $x^{-n} \neq e$ for any $n$ because we can easily reverse engineer a statement about the order of $x$ or $x^{-1}$ from the other. Suppose that $|x|$ is finite and call it $m$. Then
$$x^m = e.$$
But, using (a),
$$(x^m)^{-1} = x^{-m}.$$
But $x^m = e$, and $ee = e$, so $(x^m)^{-1} = e$. That is,
$$x^{-m} = e.$$
But we can rewrite the LHS:
$$(x^{-1})^m = e.$$
Hence, $|x^{-1}| = m = |x|$.
Some help, especially on (a) and (b), would be appreciated.
 A: It is possible to prove (a) independently. Some if not most textbooks have the standard notation (see Dummit and Foote for example) of
\begin{equation*}
a^{-n}=\underbrace{a^{-1}a^{-1}\cdots a^{-1}}_{n\text{ times}}
\end{equation*}
So with the above notation and first assuming $n>0$,
\begin{equation*}
a^{n}a^{-n}=\underbrace{aa\cdots a}_{n\text{ times}}\cdot\underbrace{a^{-1}a^{-1}\cdots a^{-1}}_{n\text{ times}}=1_{G}
\end{equation*}
where the last equality follows from the generalised associative law for groups. Therefore, $a^{-n}=(a^{n})^{-1}$. If $n<0$, then write $m=-n>0$ and use a similar argument as above. 
To prove (b), let us consider (for simplicity sake) only the case when $b<0<a$ and $|b|<a$ in $\mathbb{Z}$. Once again, with the above notation and using the generalised associative law for groups,
\begin{equation*}
x^{a}x^{b}=x^{a}x^{-|b|}=\underbrace{xx\cdots x}_{a\text{ times}}\cdot\underbrace{x^{-1}x^{-1}\cdots x^{-1}}_{|b|\text{ times}}=x^{a-|b|}=x^{a+b}.
\end{equation*}
The other cases for (b) on $x^{a+b}=x^{a}x^{b}$ is done in a similar manner. To prove $(x^{a})^{b}=x^{ab}$, we once again consider only the case when $a>0$, $b<0$. In this case it does not matter if $|b|<a$ or $|b|>a$. Using the notation as above again, and using the result in (a) and the first part of (b),
\begin{equation*}
(x^{a})^{b}=(x^{a})^{-|b|}=((x^{a})^{-1})^{|b|}=(x^{-a})^{|b|}=\underbrace{x^{-a}x^{-a}\cdots x^{-a}}_{|b|\text{ times}}=x^{-a|b|}=x^{ab}.
\end{equation*}
The case when $a<0,b<0$ is done in a similar manner.
