Prove that $3^{2n} +7$ is divisible by 8 
Prove by induction that $3^{2n} +7$ is divisible by 8 for $n \in \Bbb N$
So I think I have completed this proof but it doesn't seem very thorough to me - is my proof valid?

If $n=1$ then $3^{2n} +7 = 16 = 2(8)$ so true when $n=1$
Assume true for $n=k$ so $$ 8\vert 3^{2k} +7$$
If $n=k+1$
$$3^{2(k+1)} +7$$
$$3^{2k+2} +7$$
If $3^{2n} +7$ is divisble by 8 then $3^{2n} +7=8A$ where $A \in \Bbb Z$, and thus $3^{2k+2} +7=8B$ where $B \in \Bbb Z$
$$3^2 \times 3^{2k}+7$$
$$3^2 \times (8A)=72A$$
$$72A =8(9A)=9B$$
So by induction $3^{2n} +7$ is divisibe by 8 $\forall n \in \Bbb N$

No, it is not correct. It sholud be done something like this: 
$$3^{2k}= 8A-7$$ so\begin{eqnarray} 3^{2k+2}+7 &=& 9\cdot 3^{2k}+7\\ &=& 9\cdot (8A-7)+7 \\ &=& 72A-56\\ &=& 8(9A-7)\end{eqnarray}
