Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$ 
Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$

Proof. So i'm going to prove this by induction. The first case when $n=1$ is trivial since:
$$9+16=25,$$ 
implying $ 5 \mid 25$.
Now we need to show is divisible when $n=k+1$. We will use this later on, but $$9^k+4^{k+1}=5k.$$
The 
n,
$$\begin{align}9^{k+1}+4^{k+2} 
&= 9 \cdot 9^k+4 \cdot 4^{k+1} \\
&= 9 \cdot 9^k+(9-5) \cdot 4^{k+1} \\
&=9 \cdot 9^k+9 \cdot 4^{k+1} -5 \cdot 4^{k+1} \\
&= 9(9^k + 4^{k+1})-5 \cdot  4^{k+1} \\
&=\underbrace{9(5k)-5 \cdot 4^{k+1}}_{\text{This is where the $5k$ comes in}} \\
&= 5(9k-4^{k+1}),\end{align}$$
thus, the original expression a multiple of $5$. 
Is my induction correct?
Edit: I see several answers that took a different approach, all is welcome it really helps me see it in a different way. Thank You!

By Induction:
$n=0$: $9^0+4^1=5$ , divisible by $5$.
Hypothesis:
Assume : $9^n+4^{ n+1}$ is divisible by $5$.
Step:
$9^{n+1}+4^{n+2} =9 \cdot 9^{n} +4 \cdot 4^{n+1}=$
$(5+4)9^n +4 \cdot 4^{n+1} =$
$ 5 \cdot 9^n +4(9^n+4^{n+1})=$
$ 5 \cdot 9^n +4\cdot 5c= 5(9^n+4c)$, 
where  $c \in \mathbb{Z^+}$, since $(9^n+4^{n+1})$ 
is by hypothesis divisible by $5$,
