Trying to find what proof works best with this question? If  $ (n+1)^2   $ is even then   $ n $    is odd
The work part isn't the issue for me it's finding what is the best and most simple proof to use on this question
My three options are Direct proof, Indirect proof(Contrapositive) or proof by Contradiction
I know n is odd if $N=2k+1$
N is even if $n=2k$
I want to say solving it by Contrapositive is the best method because you flip n to Even which will make it easier
I would approach this by proving the contrapostive, as you suggest.  The contrapostive of the statement, which is equivalent to the statement, is:

If $n$ is even, then $(n+1)^2$ is odd.

Assume $n$ is even.  So $n= 2k$, for some integer $k$.  Then $(n+1)^2 = n^2+ 2n + 1 =  (2k)^2 + 2(2k) + 1 =  4k^2 + 4k + 1= 2(2k^2+ 2k) +1 = 2m+1$,  which is an odd integer, (noting that we can represent $m = 2k^2+2k$, another integer, given that $k$ is an integer.)
Direct requires proving that if $(n+1)^2$ is even then $n+1$ is even.  And the only real way I see to do that is be contrapositive.  If $n+1=2k$ is even then $(n+1)^2 = 4k^2$ is even but if $(n+1) =2k + 1$ is odd then $(n+1)^2=4k^2 + 4k + 1= 2(2k^2 + 2k) + 1$ is odd.  Or by claiming by Euclid's Lemma that $2$ is prime so if $2|(n+1)^2$ then $2|n+1$ and $n+1$ is even.  Then once we know $n+1$ is even then $n+1 = 2k$ so $n = 2k-1$ is odd.
That's not hard but it's more work.
COntrapositive is fairly.... er, dirct.....  If $n=2k$ is even then $n+1 = 2k+1$ is odd, and $(n+1)^2 = 4k^2+4k + 1$ is odd.  So contrapositively if $(n+1)^2$ is not odd then $n$ is not even.  Done.
Proof by contradiction sets up unnecessary conditions:
Suppose $(n+1)^2$ even and $n$ is even.  Then $n+1$ is odd and $(n+1)^2$ is odd so that's a contradcition.
Fine, but why did we need to suppose $(n+1)^2$ was even.
I'd go with contrapositive.
But they are all relatively easy and straightforward.