The problem above is easy to see with $n = 1$, because then every element of $X$ maps to itself and the function $f$ is obviously bijective. By $n = 2$ we have for every $x \in X$
one $y \in X$ s.t. $(x) = y, f(y) = x$. So every element of $X$ maps to exactly one other element of $X$ and $f$ is thus bijective.

However, the proof does not cover all $n \geq 1$, but only the first two $n's$. Is there a better possiblility to make a proof that covers all $n \geq 1$?