Primitive of Power of x by Exponential of a x

Theorem
Let $n$ be a positive integer.

Then:
 * $\displaystyle \int x^n e^{a x} \ \mathrm d x = \frac {e^{a x} } a \sum_{k \mathop = 0}^n \left({\left({-1}\right)^k \frac {n^{\underline k} x^{n - k} } {a^k} }\right) + C$

where $n^{\underline k}$ denotes the $k$th falling factorial power of $n$.

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \int x^n e^{a x} \ \mathrm d x = \frac {e^{a x} } a \sum_{k \mathop = 0}^n \left({\left({-1}\right)^k \frac {n^{\underline k} x^{n - k} } {a^k} }\right) + C$

$P \left({0}\right)$ is true, as from Primitive of $e^{a x}$:
 * $\displaystyle \int e^{a x} \ \mathrm d x = \frac {e^{a x} } a$

Basis for the Induction
$P \left({1}\right)$ is the case:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({r}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({r+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \int x^r e^{a x} \ \mathrm d x = \frac {e^{a x} } a \sum_{k \mathop = 0}^r \left({\left({-1}\right)^k \frac {r^{\underline k} x^{r - k} } {a^k} }\right) + C$

Then we need to show:
 * $\displaystyle \int x^{r+1} e^{a x} \ \mathrm d x = \frac {e^{a x} } a \sum_{k \mathop = 0}^{r+1} \left({\left({-1}\right)^k \frac {\left({r+1}\right)^{\underline k} x^{r + 1 - k} } {a^k} }\right) + C$

Induction Step
This is our induction step:

So $P \left({r}\right) \implies P \left({r+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \int x^n e^{a x} \ \mathrm d x = \frac {e^{a x} } a \sum_{k \mathop = 0}^n \left({\left({-1}\right)^k \frac {n^{\underline k} x^{n - k} } {a^k} }\right) + C$