Taking derivative below an integral

I am trying to solve the following question:

If $t>0$, then \begin{align*} \int_{0}^{+\infty} e^{-tx} \; dx = \frac{1}{t} \end{align*} Moreover, if $t \geq a > 0$, then $e^{-tx} \leq e^{ax}$. Use this and Exercise 4.M. to justify differentiating under the integral sign and to obtain the formula \begin{align*} \int_{0}^{+\infty} x^n e^{-x} \; dx = n! \end{align*}

$\bf{Note:}$ Exercise 4.M. states that if $X = [0, \infty)$ and $\lambda$ is the Lebesgue measure and $f$ is a non-negative function on X, then
$$\int f \; d\lambda = \lim_{b\to \infty} \int_0^b f \; d\lambda$$

I would like to use a Corollary of the Dominated Convergence Theorem, $$\frac{d}{dt}\int f(x,t) \; d\mu(x) = \int \frac{\partial f}{\partial t}(x,t) \; d\mu(x)$$

However, for in order to use this I need to find an integrable function $g$ such that $$\left|\frac{\partial f}{\partial t}(x,t)\right| \leq g(x)$$

I guess I am having difficulty finding this $g(x)$. If I start with $f(x,t) = e^{-tx}$ then, $$\left|\frac{\partial f}{\partial t}(x,t)\right| \leq xe^{-tx} \leq xe^{ax}$$

How would I show that this last term is in $L_1$? Moreover, I assume that I would have to iterate this process several times so what I am trying to bound should actually be $x^n e^{-tx}$. If so, how should I go about constructing my $g$?

Also if anyone could redirect me to problems of this similar structure I would appreciate it, as I find them rather interesting. I am currently using Bartle's "The Elements of Integration and Lebesgue Measure".

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If $t \geq a$ then $-t \leq -a$ and the dominating function $\phi(x)=xe^{-ax}$ is in $L^1([0,\infty))$ for any $a>0$. –  беркай Apr 17 '12 at 9:27
I came up with a bound: \begin{align*} e^{ax/2} = 1 + \frac{a}{2}x &+ \frac{a^2}{2^2 2!}x^2 + \frac{a^n}{2^n n!}x^n + \ldots \intertext{thus,} x^n &\leq \frac{2^n n!}{a^n}e^{ax/2} \\ x^n e^{-ax} &\leq e^{-ax} \left(\frac{2^n n!}{a^n}e^{ax/2} \right)\\ &= \frac{2^n n!}{a^n}e^{-ax/2}. \end{align*} –  katari Apr 23 '12 at 3:42