Prove by induction $7^n$ is an odd number for every natural number n I have proved the base case already of n=1 where $7 = 2p+1$
Then I assumed $n=k$ for $7^k = 2p+1$ for $k \in N$ and  $p \in N$
To prove $7^{k+1} = 2p+1$  I have these steps so far:
$7(7^k) = 2p+1$
$7(2p+1) = 2p+1$
$14p+7 = 2p+1$
I am not sure how to prove from here.

Just rewrite your sum as
$$14p+7=2(7p+3)+1$$
Also, it's not a good idea to use $p$ for all of your steps.  If $7^k=2p+1$, then you need to show that $7^{k+1}=2q+1$ for some integer $q$.  Otherwise the equality is not actually equal since you are using the same expression as your desired form for different valued expressions.
Now since $14p+7=2(7p+3)+1$, let $q=7p+3$.  Thus $7^{k+1}=2q+1$. And you're done.
