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!

Or another way:
$$9^n + 4^{n+1} = (10-1)^n +(5-1)^{n+1}=10c+(-1)^n + 5d+(-1)^{n+1}=10c+5d = 5e$$
Edit:
Another similar way:
$$9^n + 4^{n+1}= 9^n+4\cdot(9-5)^n = 9^n+4(9^n+5c)= (1+4)9^n + 20c = 5e$$
