For your first question:
They are using the fact that if $f(x)< g(x)$ on $(a,\infty)$, then $\int_a^\infty f(x)\,dx<\int_a^\infty g(x)\, dx$. This is a standard comparison test for improper integrals.
Here, we have $y^{n+1}>x^{n+1}$ for $y$ in the interval $(x,\infty)$ (note, then, that $y> x$); so for $y$ in the interval $(x,\infty)$, we have ${e^{-y}\over y^{n+1}} <{e^{-y}\over x^{n+1}}$. Thus $\int_x^\infty {e^{-y}\over y^{n+1}} \,dy<\int_x^\infty {e^{-y}\over x^{n+1}}\, dy$.
Finally, since the integration is with respect to $y$, the term $1\over x^{n+1}$ is a constant as far as the integration is concerned and can be factored out of the integral sign.
Though it would lead to the correct result, you shouldn't think of pulling $y^{n+1}$ out first, since you are integrating with respect to $y$. You can change it to $x^{n+1}$ first, introducing an inequality, and then pull it out.