Question: How can I solve this indefinite integral for an arbitrary integer $n>0$?

$$ \int{x^n e^x dx}$$

I could partially integrate it for small $n$, but that's not really a solution.

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**Edit:** (TB) This question is closely related to: [Is there a closed form solution for $\int x^n e^{cx}$?](https://math.stackexchange.com/questions/21516/is-there-a-closed-form-solution-for-int-xn-ecx), but it is more elementary, because $n$ is an integer here.

Answer: 
Hint: Use integration by parts.

EDIT: Try several values of $n$.
$$
\int {x e^x dx} = (x - 1)e^x + C
$$
$$
\int {x^2 e^x dx} = (x^2 - 2x + 2)e^x + C.
$$
$$
\int {x^3 e^x dx} = (x^3 - 3x^2 + 6x - 6)e^x + C.
$$
$$
\int {x^4 e^x dx} = (x^4 - 4x^3 + 12x^2 - 24x + 24)e^x + C.
$$
$$
\int {x^5 e^x dx} = (x^5 - 5x^4 + 20x^3 - 60x^2 + 120x - 120)e^x + C.
$$
Conclude that
$$
\int {x^n e^x dx} = \bigg[\sum\limits\_{k = 0}^n {( - 1)^{n - k} \frac{{n!}}{{k!}}x^k } \bigg]e^x + C.
$$
You could use the generating function approach.  

$$ \eqalign{\int\_0^X e^{tx} e^x \ dx &= \frac{e^{(1+t)X} - 1}{1+t}\cr 
&= \sum\_{k=0}^\infty (-1)^k t^k \left(e^X -1 + \sum\_{j=1}^\infty e^X \,\frac{X^j}{j!} t^j\right)\cr
 &= \sum\_{n=1}^\infty \left((-1)^n (e^X - 1) + \sum\_{j=1}^n (-1)^{n-j} \frac{X^j}{j!} e^X \right) t^n\cr}$$
But also 
$$ \int\_0^X e^{tx} e^x \, dx = \sum\_{n=0}^\infty \frac{t^n}{n!} \int\_0^X x^n e^x \, dx$$
Equating coefficients of $t^n$ from both sides,
$$ \int\_0^X x^n e^x\, dx = (-1)^n n! (e^X - 1) +
\sum\_{j=1}^n (-1)^{n-j} \frac{n!}{j!} X^j e^X $$