How do you show that $e^y = \displaystyle \int_{0}^{\infty} \dfrac{y^n} {n!}~dy$? and yet, my textbook yields the following equation from that one, $n! = \displaystyle \int_{0}^{\infty} y^ne^{-y}~dy$
Here is my attempt: $\displaystyle n!~\lim_{a \to \infty} \int_{0}^{a} y^n~dy \\ = n!~\lim_{a \to \infty \left[\dfrac{1}{n+1} y^{n+1}\right]_0^a \\ = \dfrac{n!} {n+1} \lim_{a \to \infty} a^{n+1}$ What is the next step?
You don't.
It is correct that $n! = \displaystyle \int_{0}^{\infty} y^ne^{-y} dy $.
You can not take the $e^{-y}$ out of the integral. In general, you can not take any function of the variable of integration ($y$ in this case) out of the integral.