Elementary Methods in Number Theory Exercise 1.5.5

Let $a$ and $n$ be positive integers.Prove that $a^n-1$ is prime only if $a=2$ and $n=p$ is prime.

Proof:
\begin{equation}
a^n-1^n=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)
\end{equation}

So
\begin{equation}
a=2
\end{equation}
So
\begin{equation}
a^n-1=2^n-1
\end{equation}
If $n$ is not prime,then $n=p_1p_2,p_1,p_2>1$.Then

\begin{equation}
2^n-1=2^{p_1p_2}-1=(2^{p_1})^{p_2}-1=(2^{p_1}-1)\Delta
\end{equation}
$\Delta>1$.So $2^n-1$ become a composite number,which leads to absurdity.So $n$ is a prime.