Compute $\int^\infty_0 t^{n-1} e^{-t} dt$. What to do with $n$? I need to find 
$$f(n) = \int^\infty_0 t^{n-1} e^{-t} dt$$
So I think I find the indefinate integral first? But what do I do with $n$, since I am integrating with respect to $t$?
 A: Here is also another approach, though a posteriori.
Tonelli's theorem enables us to exchange the order of integration and summation of a sequence of nonnegative functions. Thus for $ 0 < r < 1 $,
$$\begin{align*} \sum_{n=0}^{\infty} \frac{r^n}{n!} \int_{0}^{\infty} x^n e^{-x} \; dx & = \int_{0}^{\infty} \sum_{n=0}^{\infty} \frac{(rx)^n}{n!} e^{-x} \; dx = \int_{0}^{\infty} e^{-(1 - r)x} \; dx \\ & = \frac{1}{1 - r} = \sum_{n=0}^{\infty} r^{n}. \end{align*}$$
Actually, it's essentially identical to cooper.hat's approach, since it is just the Taylor expansion of $f(1-r)$ with his/her $f$.
A: This integral can be easily computed by means of integration by parts. Let's call it $$\Gamma(n) =\int\limits_0^\infty t^{n-1}e^{-t}dt$$
We want to show that $\Gamma(n)$ follows some law. 
Let's integrate by parts, with $u=t^{n-1}$. Then 
$$\int\limits_0^\infty t^{n-1}e^{-t}dt=\left.-e^{-t}t^{n-1}\right|_0^\infty+(n-1)\int\limits_0^\infty t^{n-2}e^{-t}dt$$
but since $\left.-e^{-t}t^{n-1}\right|_0^\infty$ vanishes, we have that
$$\int\limits_0^\infty t^{n-1}e^{-t}dt=(n-1)\int\limits_0^\infty t^{n-2}e^{-t}dt$$ or that
$$\Gamma(n)=(n-1)\Gamma(n-1)$$
We can go on and use the relation to get
$$\Gamma(n)=(n-1)!$$
Another approach is copper's:
Define $$I(x) = \int\limits_0^\infty e^{-tx} dt $$
By means of Leibniz' rule, we have that
$$I^{(n)}(x) = \int\limits_0^\infty (-t)^ne^{-tx} dt $$
So we're interested in
$$(-1)^nI^{(n)}(1) = \int\limits_0^\infty t^n e^{-t} dt $$
But since
$$I(x) = \int\limits_0^\infty e^{-tx} dt=\frac{1}{x}$$ and we can simply show that
$$I^{(n)}(x)=\frac{(-1)^n n!}{x^{n+1}}$$ it follows that
$$(-1)^nI^{(n)}(1)=\frac{ n!}{1^{n+1}}=n!$$ as desired.
A: If $n>0$ is an integer, here is another approach: Let $I(x) = \int^\infty_0 e^{-x t} dt$, with $x>0$. It is straightforward to evaluate $I(x) = \frac{1}{x}$, and notice that $f(1) = I(1) = 1$. To continue, notice that $\frac{d I(x)} {d x} = \int^\infty_0 (-t) e^{-x t} dt$, and by direct computation, $\frac{d I(x)} {d x} = -\frac{1}{x^2}$, so $f(2) = -\frac{d I(1)} {d x} = 1$. The process may be continued by induction.
A: $$\int^\infty_0 t^{n-1} e^{-t} dt=\Gamma(n)$$
A: I think $n$ is parametr. So you have to do it, treating $n$ like normal number.
