Checking whether a polynomial of high degree is bijective or not.

$P(x) = x^{101}$ is a bijective polynomial of degree $101$. $Q(x) = x^{101}-x^2$ is of degree $101$, too, but it's not injective. So unless there's more to the question, an arbitrary 101st degree polynomial can be bijective, but not all of them are.

A polynomial $P:\mathbb{R}\to\mathbb{R}$ is bijective if and only if $P'(x)$ never changes sign.

In less mathematical language, we need the polynomial to always go up, or always go down, and it's only allowed to level out momentarily.

With this in mind, it is fairly easy to write down a polynomial of odd degree that is bijective: only allow positive coefficients and terms of odd degree. For example, consider $P(x)=x^{101}+3x^{35}+4x$. You will find that its derivative is always positive.

It is not true that all bijective polynomials have the above form, as you can also check that $P(x)=x^{101}-x^2+10x$ also has an everywhere positive derivative.

Although (as Javier demonstrated) not all polynomials of odd degree are injective, they are all surjective. Say $p(x) = a_n x^n + \ldots + a_0$, where $n$ is odd. Then if $a_n > 0$, we will have $\lim_{x\to+\infty} p(x) = +\infty$ and $\lim_{x\to-\infty} p(x) = -\infty$. Since polynomials are continuous, we have to have hit everything in between somewhere along the way, so $p(x)$ is surjective (if $a_n < 0$, then $\lim_{x\to+\infty} p(x) = -\infty$ and $\lim_{x\to-\infty} p(x) = +\infty$). This happens because as $\left|x\right|$ becomes large, the highest degree term starts to contribute more and more compared to the other terms, so eventually it overwhelms them and the polynomial will simply take the sign of $a_n x^n$.