integration - Integrating this improper integral to test for convergence?

I'm trying to integrate this:

$$\int^\infty_0 \frac{8}{\sqrt{e^{x}-x}} \,dx$$

And use the Direct Comparison Test to find out whether it diverges or converges.

I looked at a similar problem:

and I can see how the integrals on the lefthand side are less than the integrals on the righthand side, since the rightmost right-side integral is squared from the rightmost left-side integral, but:

Why is the 5 dropped? Is it because a small added constant ultimately wouldn't affect the behavior of x on its way to infinity?

Why is the integral from 1 to infinity squared, out of all the possible operations we could perform on it?

And is this the correct next step in my own integration?

$$\int^\infty_0 \frac{8}{\sqrt{e^{x}-x}} \,dx = \int^1_0 \frac{8}{\sqrt{e^{x}-x}} \,dx + \int^\infty_1 \frac{8}{\sqrt{e^{x}-x}} \,dx < \int^1_0 \frac{8}{\sqrt{e^{x}-x}} \,dx + \int^\infty_1 \frac{1}{\sqrt{e^{x}}} \,dx$$

Thank you in advance if you're able to help clarify this.

Answer

Note that $e^x-x \geq x^4$ for all sufficiently large $x$. So there exists some $N > 0$ such that $e^x-x \geq x^4$ for all $x \geq N$. Since

$$\sqrt{e^x-x} \geq \sqrt{x^4} = x^2 \quad \Rightarrow \quad \frac{1}{\sqrt{e^x-x}} \leq \frac{1}{x^2}$$ for all $x \geq N$, we have $$\begin{align*} \int_0^\infty \frac{dx}{\sqrt{e^x-x}} & = \int_0^N \frac{dx}{\sqrt{e^x-x}} + \int_N^\infty \frac{dx}{\sqrt{e^x-x}} \\ & \leq \int_0^N \frac{dx}{\sqrt{e^x-x}} + \int_N^\infty \frac{dx}{x^2}. \end{align*}$$ The two integrals are finite so the integral you consider is convergent.