$9^n \equiv 1 \mod 8$ I would like someone to check this inductive proof (sketch)
The base case is clear.  For the inductive step, it follows that $8 \mid 9^{n+1} - 9 = 9(9^n - 1)$ by the indutive hyp.  So $9^{n+1} \equiv 9 \equiv 1 \mod 8$.
Feedback would be appreciated.

I'm assuming you mean what you say when you state your work as a proof "sketch".

The base case is clear.  For the inductive step, it follows that $8 \mid 9^{n+1} - 9 = 9(9^n - 1)$ by the indutive hyp.  So $9^{n+1} \equiv 9 \equiv 1 \mod 8$.

In your final write up, I'd suggest you "fill in" a bit of detail: e.g., to "walk through" the base case, at least stating that the congruence holds for $n=1$, or perhaps
"for $n = 1$, clearly, $9\equiv 1 \pmod 8$". 
Then I suggest you make your inductive hypothesis explicit: 
"Assume that it is true that $9^n \equiv 1 \pmod 8$,"
and then finish with, "for the inductive step....[what you wrote]"

If your task was to prove the congruence holds using proof by induction on $n$, then you've done a fine job of sketching such a proof.
If you can use other strategies, then bonsoon's suggestion is worth considering:
"Or note that since $9 \equiv 1 \pmod 8$, we have $9^n\equiv 1^n = 1 \pmod 8.$"
