$9^n \equiv 1 \mod 8$ 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.

Yes, that is correct. Alternatively, prove by induction the $\,\rm n$-ary congruence product rule
$$\rm\qquad\ \ a_k\equiv b_k\ \Rightarrow\ a_1\cdots\, a_n \equiv b_1\cdots\, b_n$$
by iterating the binary  product rule  $\rm\ a_k\equiv b_k\ \Rightarrow\ a_1 a_2 \equiv b_1 b_2,\:$  then specialize $\rm\:a_i \equiv 9,\ b_i\equiv 1$
Remark $\ $ Your proof can be viewed as special case of the obvious inductive proof that a sequence $\rm\:f_n\:$ is constant if successive values never change, i.e. if  $\rm\:f_{n+1} \equiv f_n.$ Indeed, in your special case we have $\rm\:mod\ 8\!:\ f_{n+1} = 9^{n+1} = 9\, f_n \equiv f_n\:$ so the sequence is constant, hence $\rm\:f_n\equiv f_0\equiv 9^0\equiv 1.\:$ 
