<s>[INST] Prove that $5^n + 6^n -1$ is divisble by 10 for all positive integers n. I know the base case is:
$$ n=1,
5^1+6^1-1=10,10|10
$$
But I am struggling about how to go about the inductive step.
 [/INST] $5^n-1$ is even implies that $5^n+6^n-1$ is divisible by $2$,
$6^n-1=(6-1)(6^{n-1}-1+..)$ ($6=1$ mod $5$ implies that $6^n=1$ mod $5$ and $6^n-1=0$ mod $5$) implies that $5^n+6^n-1$ is divisible by $5$.
Recursive proof
$5^{n+1}+6^{n+1}-1=5.5^n+(5+1)6^n-1=5(5^n+6^n-1)+6^n+4$
Suppose that $6^n=6$ mod $10$, $6^{n+1}=36=6 $ mod $10$   implies that $6^n+4=0$ mod $10$.
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