Why is $5^{n+1}+2\cdot 3^n+1$ a multiple of $8$ for every natural number $n$? I have to show by induction that this function is a multiple of 8. I have tried everything but I can only show that is multiple of 4, some hints? The function is
$$5^{n+1}+2\cdot 3^n+1 \hspace{1cm}\forall n\ge 0$$, because it is a multiple of 8, you can say that$$5^{n+1}+2\cdot 3^n+1=8\cdot m \hspace{1cm}\forall m\in\mathbb{N}$$.

By induction:
It is true for case $n=1$, Let it be true for $n=k$ then 
$5^{k+2} +2.3^{k+1} +1 = 5.5^{k+1}+3.2.3^k+1 = 2(5^{k+1} -1) + 3(5^{k+1}+2.3^k+1)$
The second part is a multiple of $8$ and one can easily  show that $(5^{k+1} -1)$ is a multiple of $4$.
Hint for showing  $(5^{k+1} -1)$ is a multiple of $4$:
$5^{k+2}-1 = 4.5^{k+1} + 5^{k+1} -1$
