Prove $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$ by induction Here is my attempted proof:
$\forall n \in \mathbb{N}$, let $S_n$ be the statement: $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$
Base case: $S_1$: $\frac{2^{4(1)}-(-1)^1}{17} = \frac{16+1}{17} = 1 \in \mathbb{N}$
Inductive step: $\forall n \geq 1$, $S_n$ holds
$S_{n+1}$: $\frac{2^{4(n+1)}-(-1)^{n+1}}{17} = \frac{2^{4n} \cdot 2^4 +(-1)^{n}}{17} = \frac{2^{4n} \cdot (2^4+1) - 2^{4n} +(-1)^{n}}{17}$.
Now, we are left with $2^{4n} - \frac{2^{4n}-(-1)^n}{17}$. 
We know $\frac{2^{4n}-(-1)^n}{17}$, $2^{4n} \in \mathbb{N}$ and $2^{4n} > \frac{2^{4n}-(-1)^n}{17} \implies 2^{4n} - \frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$
$$\tag*{$\blacksquare$}$$ 
Is my proof correct? I also feel there is a more elegant/smoother proof (assuming mine is correct).

Alternatively, assuming $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$:
$$\begin{align}\frac{2^{4(n+1)}-(-1)^{n+1}}{17} &= \frac{2^{4n} \cdot 2^4 +(-1)^{n}}{17} =\\
&=\frac{2^{4} \cdot (2^{4n}-(-1)^n) + 2^{4}(-1)^n +(-1)^{n}}{17}=\\
&=2^4\cdot \underbrace{\frac{2^{4n}-(-1)^n}{17}}_{\ge 1}+(-1)^{n}\in \mathbb N.
\end{align}$$
