As already noted, the answer is "yes". Here is one way to see it: factor out the leading term, so the polynomial has the form $P(x) = a x^n( 1 + \text{ expression in powers of } \frac{1}{x}).$
If $x$ is very positive or very negative, i.e. if $|x| \gg 0$, then the expression in powers of $\frac{1}{x}$ will be negligible, and so $P(x) \sim a x^n$ for $| x | \gg 0.$
So now just look at the behaviour of $a x^n$. If $n$ is odd this is very negative when $x$ is very negative, becoming more so as $x\to -\infty$, while it is very positive when $x$ is very positive, becoming more so as $x \to \infty$. If $n$ is even then it is very positive when $x$ is either very negative or very positive, and increases both as $x \to -\infty$ and as $x \to \infty$.
Edit: As Didier Piau notices, this argument with crude asymptotics is not precise enough to conclude true monotonicity for $|x| \gg 0$; Moron's argument with derivatives is better for that. However, it does give an explanation for the rough behaviour of $P(x)$ for $|x| \gg 0$.