TITLE: Computing the limit $\lim_{k \to \infty} \int_0^k x^n \left(1 - \frac{x}{k} \right)^k \mathrm{d} x$ for fixed $n \in \mathbb{N}$
QUESTION [11 upvotes]: I'm working on a problem that asks to compute
$$\lim_{k \to \infty} \int_0^k x^n \left(1 - \frac{x}{k} \right)^k \mathrm{d} x$$
for fixed $n \in \mathbb{N}$. What I've tried so far is to do a $u$-substitution for $u = \frac{x}{k}$, so I have
$$\int_0^k x^n \left(1 - \frac{x}{k} \right)^k \mathrm{d} x = k^{n + 1} \int_0^1 u^n (1 - u)^k \mathrm{d} u .$$
Using the Binomial Theorem to break up the $(1 - u)^k$ term, I get
$$k^{n + 1} \int_0^1 u^n (1 - u)^k \mathrm{d} u = k^{n + 1} \int_0^1 \sum_{j = 0}^k \binom{k}{j} \frac{(-1)^j}{n + j + 1} .$$
However, I don't know how to compute the limit of this expression as $k \to \infty$. I assume that I should recognize it as some kind of Taylor series that's somehow $O \left( k^{-(n + 1)} \right)$, but I'm not seeing it.

Note: When looking at other posts, I found an integral that looked similar to this one, and the only answer on that post involved something called a beta function. I have never heard of a beta function, and would like to find a solution here that doesn't rely on whatever a beta function is.

Another idea I considered was to use the Dominated Convergence Theorem, since $e^{-x} = \lim_{k \to \infty} \left( 1 - \frac{x}{k} \right)^k$, so I figured I could use DCT to say that
\begin{align*}
\lim_{k \to \infty} \int_0^k x^n \left( 1 - \frac{x}{k} \right)^k \mathrm{d} x & = \lim_{k \to \infty} \int_0^\infty \chi_{[0, k]}(x) x^n \left( 1 - \frac{x}{k} \right)^k \mathrm{d} x \\
& = \int_0^\infty x^n e^{-x} \mathrm{d} x & (\textrm{DCT used here})\\
& = n ! ,
\end{align*}
assuming I didn't mess up any of my integration by parts. However, I couldn't find a choice of dominator that would work on all of $[0, \infty)$, so I'd also be interested in a solution that uses DCT as well. Perhaps I'm being naive, but the pointwise limit is just so convenient that I have to imagine that DCT can be used here.
EDIT: Thanks to some inspiration from a comment by user Mars Plastic, I thought to consider some other integral convergence theorems. I came up with this, which I think works. I'm still interested to see if the Dominated Convergence Theorem argument can be made to work, perhaps a bit more smoothly than this.
Let $f_k(x) = \chi_{[0, k]}(x) x^n \left( 1 - \frac{x}{k} \right)^k$. I claim that the sequence is monotone increasing on $[0, \infty)$. Fix $x \in (0, \infty)$, and let $K = \lfloor k \rfloor + 1$, so that $K = \min \{ k \in \mathbb{N} : f_k(x) \neq 0 \}$. Obviously if $k < K$, then $f_k(x) = 0 \leq f_{k + 1}(x)$. So consider the case where $k \geq K$. Then $f_k(x) = x^n \left(1 - \frac{x}{k} \right)^k$, and based on answers to this question, it seems this sequence would be monotone increasing. Therefore, I can apply the Monotone Convergence Theorem to say that $f_k(x) \nearrow x^n e^{-x}$, so $$\int_0^k x^n \left( 1 - \frac{x}{k} \right)^k \mathrm{d} x = \int_0^\infty f_k(x) \mathrm{d} x = \int_0^\infty x^n e^{-x} = n! .$$

REPLY [3 votes]: Let $\displaystyle I(n, k) =  \int_0^1 u^n (1 - u)^k \,\mathrm{d} u$. Then integrating by parts:
$\begin{aligned}  I(n, k) & =  \frac{1}{n+1}\int_0^1 (u^{n+1})' (1 - u)^k \,\mathrm{d} u \\& = \frac{1}{n+1}\cdot u^{n+1}(1-u)^{k}\bigg|_0^1-\frac{1}{n+1}\int_0^1 u^{n+1}[(1-u)^{k}]'\, \mathrm du \\& = \frac{k}{n+1}\int_0^1 u^{n+1}(1-u)^{k-1}\, \mathrm du  \\& = \frac{k}{n+1} I(n+1, k-1).  \end{aligned}$
Carrying out the recurrences:
$\begin{aligned}  I(n, k) & = \frac{k}{n+1} I(n+1, k-1) \\& = \frac{k}{n+1}\cdot \frac{k-1}{n+2} \cdots \frac{1}{n+k}I(n+k, 0) \\& = \frac{k}{n+1}\cdot \frac{k-1}{n+2} \cdots \frac{1}{n+k}\int_0^{1} u^{n+k} \, \mathrm du 
 \\& = \frac{k}{n+1}\cdot \frac{k-1}{n+2} \cdots \frac{1}{n+k} \cdot \frac{1}{n+k+1} \\& = \frac{k!n!}{(k+n+1)!}. \end{aligned}  $
Therefore
$$I = \lim_{k \to \infty} \frac{k!n! k^{n+1}}{(k+n+1)!} = n! \cdot \lim_{k \to \infty} \frac{k! k^{n+1}}{(k+n+1)!} = n!$$