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.

If $7^k$ is odd i.e. $7^k=2p+1$ for some integer $p$, then
$$
7^{k+1}=7(2p+1)=14p+7=2(7p+3)+1
$$
as desired.
