Elementary Methods in Number Theory Exercise 1.5.6

Let $k$ be a positive integer,prove that if $2^k+1$ is prime,then $k=2^n$.

Proof:If $\forall n$,$k\neq 2^n$,then there exists a prime number $p\neq 2$ such that $k=pt$. Then
\begin{equation}
2^k+1=(2^t)^p+1
\end{equation}
$p$ must be an odd number,then
\begin{equation}
(2^t)^p+1=(2^t+1)\Delta
\end{equation}
$\Delta>1$.So $2^k+1$ become a composite number,which leads to absurdity.So $k=2^n$.