My attempt:
Let's suppose there is an even number $n$ such that $n^4$ is odd. Then, since $n$ is even, $n= 2k$ where $k$ is an integer. Then, $n^4 = 16k^4 = 2(8k^4)$ which is an even number. Therefore, there's no even $n$ such that $n^4$ is odd.
Is this proof correct ?
You can prove the statement using the contrapositive, which is an equivalent statement, of the original.
Contrapositive: If n is even, then $n^4$ must be even.
We can now build a contradiction from this statement by assuming n is odd.
Let n be odd, then there exists an integer k such that $n=2k+1$.
$n^4=(2k+1)^4=16k^4+32k^3+24k^2+8k+1$
$=2(8k^4+16k^3+12k^2+4k)+1$ which is odd.
Therefore, you have a contradiction to your contrapositive statement. QED