Primitive of Power of x by Exponential of a x

Theorem
Let $n$ be a positive integer.

Let $a$ be a non-zero real number.

Then:

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

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
 * $\ds \int x^n e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} x^{n - k} } {a^k} } + C$

$\map P 0$ is true, as from Primitive of $e^{a x}$:
 * $\ds \int e^{a x} \rd x = \frac {e^{a x} } a$

Basis for the Induction
$\map P 1$ is the case:

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\ds \int x^r e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^r \paren {\paren {-1}^k \frac {r^{\underline k} x^{r - k} } {a^k} } + C$

Then we need to show:
 * $\ds \int x^{r + 1} e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^{r + 1} \paren {\paren {-1}^k \frac {\paren {r + 1}^{\underline k} x^{r + 1 - k} } {a^k} } + C$

Induction Step
This is our induction step:

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

Therefore:
 * $\ds \int x^n e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} x^{n - k} } {a^k} } + C$