Show that $\int_{0}^{\infty}x^ne^{-x}\,dx=n!$ by differentiating the equality $\int_{0}^{\infty}e^{-tx}\,dx=\frac{1}{t}$ I'm learning about measure theory, specifically Lebesgue integral, and need help with the following problem:

Show that $\int_{0}^{\infty}x^ne^{-x}\,dx=n!$ by differentiating the equality $\int_{0}^{\infty}e^{-tx}\,dx=\frac{1}{t}$.

Since I had no idea how to solve this problem I first did some research and come across the definition of the Gamma function which I think this problem is related to. I have found that
$$\Gamma(x+1)=\int_{0}^{\infty}t^xe^{-t}\,dt.$$
Using integration by parts it is actually quiet easy to show that the RSH improper integral evaluates to $$\Gamma(x+1)=x\Gamma(x).$$ 
From the above result, substituting successively $x$ with $n-1, n-2,\ldots$ we get that the Gamma function is indeed the factorial function. In particular
$$\Gamma(n+1)=n!$$

My issue is that the problem specifically says to use the fact that $\int_{0}^{\infty}e^{-tx}\,dx=\frac{1}{t}$ and differentiate it. Nowhere in the above prove am I using this equality. How do I show that the improper integral is equal to the factorial function using the given equality? 
 A: Let $f(t)$ be the function represented by the integral
$$\begin{align}
f(t)&=\int_0^\infty e^{-tx}\,dx\\\\
&=\frac1t
\end{align}$$
Note that the $n$'th derivative, $f^{(n)}(t)$, of $f(t)$ is given by
$$\begin{align}
f^{(n)}(t)&=\int_0^\infty (-1)^n x^n\,e^{-tx}\,dx\\\\
&=(-1)^n\,n! \frac{1}{t^{n+1}} \tag 2
\end{align}$$
Dividing both sides of $(2)$ by $(-1)^n$ and setting $t=1$ we find that
$$\int_0^\infty x^ne^{-x}\,dx=n!$$
as was to be shown!


NOTE:


Here, we justify interchanging the derivatives with the integral.  Note that we have


$$\frac{f(t+h)-f(t)}{h}=\int_0^\infty \frac{e^{-hx}-1}{h} e^{-tx}\,dx$$


For $|h|\le t/2$, $\left|\frac{e^{-hx}-1}{h}\right|\le \max\left(\frac2t e^{tx/2},x\right)$ for all $x\ge 0$.  Therefore, the integrand is bounded as


$$\left|\frac{e^{-hx}-1}{h} e^{-tx}\right|\le \max\left(xe^{-tx},\frac2t e^{-tx/2}\right)$$


Inasmuch as $\int_0^\infty \max\left(xe^{-tx},\frac2t e^{-tx/2}\right)\,dx<\infty$, the Dominated Convergence Theorem guarantees that


$$\begin{align}
\lim_{h\to 0}\int_0^\infty \frac{e^{-hx}-1}{h} e^{-tx}\,dx&=\int_0^\infty \lim_{h\to 0}\frac{e^{-hx}-1}{h} e^{-tx}\,dx\\\\
&=-\int_0^\infty xe^{-xt}\,dx
\end{align}$$


The same argument is applicable for taking the second and higher-ordered derivatives.

A: 
Show that $\int_{0}^{\infty}x^ne^{-x}\,dx=n!$

Without using the requirement, just noting this is a classic "integration by parts" result.
$$
\begin{align}
\int x^n e^{-x}
&=x^n \int e^{-x}-\int nx^{n-1}(\int e^{-x}dx)dx\\
&=-x^ne^{-x}+n\int x^{n-1}e^{-x}dx
\end{align}
$$
and the limits $0$ and $\infty$ make the $x^n e^{-x}$ term disappear.
