These kinds of proofs never feel rigorous to me, so I might be missing something. Here's what I did:
For all $M<0$, if $x<0$,
$x^n > 0 > M$
For all $M\ge0$, if $x<-\root{n}\of{M}$,
$x^n > M$
And that's it. To me, this does not feel right. Do I need to expand more on proving that $x^n > 0$ for the first part, or maybe that $-\root{n}\of{M}$ even exists, or that $x<-\root{n}\of{M}$ results in $x^n > M$?
I have this feeling for any proof I do regarding "if $n$ is even/odd" where $n$ is a power. What am I missing?
I would definately go for the route of proving that $x^n$ for $x<0$ and for positive even integer $n$, results always being positive. This is a stepping stone for showing the "unboundedness" itself, and lends for easier solutions to show that the function of $x^n$ is not bounded by any given $M$, from which you can easily conclude that $x^n$ really diverges.
To be a bit blunt, you are now only proving that $x^n$ is only greater than some $M$ given that some conditions are met, for the relationship of $x$ and $M$. Tautologous, I would say.
What is missing, is that you have tie the strings to show that given any $M$, there always exists a point(s) of $x^n$ that are larger than the given $M$. And for this purpose, I would prove that the $x^n$ is always positive with the given conditions, and it's also clearly continuous and increasing, which grants a possibility to choose sufficient $x_M$ to get $x_M^n>M$.
In my opinion going through even and odd $n$ is unnecessary, unless that is a way you want to show the positiveness of the $x^n$.