Prove $\lim\limits_{n \to \infty} \int_0^1 x^n e^x dx = 0$ by evaluating closed form While preparing for my numerical analysis exam I encountered the following question

Find a recursive definition $I_n = R(I_{n - 1})$ for $I_n := \int_{0}^{1} x^n e^x dx$ and prove
  $$
I_n = \sum_{k = 0}^{n} (-1)^{k} \frac{n! e}{(n - k)!} - (-1)^n n!
$$
  and give an explanation for the behaviour of $I_n$ for $n \to \infty$.

With integration by parts I found $I_n = e - n I_{n - 1}$, $I_0 = e - 1$ and proved the formula with induction.
Just for reference, is the following induction step correct?
\begin{align}
I_{n + 1} = e - (n + 1)I_n
& = e - (n + 1)\left( \sum_{k = 0}^{n} (-1)^{k} \frac{n! e}{(n - k)!} - (-1)^n n!\right) \\
& = e - \sum_{k = 0}^{n} (-1)^{k} \frac{(n + 1)! e}{(n - k)!} - (-1)^{n + 1} (n + 1)! \\
& = e\left( 1 - \sum_{k = 0}^{n} (-1)^{k} \frac{(n + 1)!}{(n - k)!} \right) - (-1)^{n + 1} (n + 1)! \\
& = e\left( 1 - \sum_{k = 1}^{n + 1} (-1)^{k - 1} \frac{(n + 1)!}{(n - k + 1)!} \right) - (-1)^{n + 1} (n + 1)! \\
& = \sum_{k = 0}^{n + 1} (-1)^{k} \frac{(n + 1)! e}{(n + 1 - k)!} - (-1)^{n + 1} (n + 1)!
\end{align}
Now I am trying to investigate the behaviour for $n \to \infty$.
First I rewrote
$$
I_n
= \sum_{k = 0}^{n} (-1)^k n! \left( \frac{e}{(n - k)!} - (-1)^{n - k}\right)
$$
Wolfram Alpha yields
$$
= (-1)^{n + 1} n! \left( n + 1 - e \sum_{k = 0}^{n} \frac{(-1)^k}{k!} \right),
$$
but I don't know how to proceed from here.
I though that the last term would cancel out since $\sum_{k = 0}^{\infty} \frac{(-1)^k}{k!} = \frac{1}{e}$ but I am not sure.
On the question paper, we are given the following numerical values:


*

*0.2280015155 for n = 10,

*0.1238038308 for n = 20,

*0.08502696925 for n = 30,

*0.06475689045 for n = 40 and

*0.05229363830 for n = 50.
They are all positive and seem to be converging to 0.
But because of the $(-1)^{n + 1}$ term I wouldn't expect this to converge.
Intuitively the result make perfect sense because for larger $n$ the $x^n e^x$ is very small for "most of" $[0,1]$ and then "in the last moment" rises steeply to $(1,1)$.
 A: Your induction step seems correct to me.
However, your approach is not best for the asymptotic behaviour. I suggest that you prove for instance that $0 \leq I_n \leq e\int_0^1{x^n\,dx}$. 
A: Note that
$$I_n = n!(-1)^ne\left(\sum_{k = 0}^{n}  \frac{(-1)^{n-k}}{(n - k)!} - e^{-1}\right)=n!(-1)^ne\left(\sum_{k = 0}^{n}  \frac{(-1)^{k}}{k!} - e^{-1}\right).$$
Now recall Taylor's theorem with the Lagrange remainder and apply it to $f(x)=e^x$ with $x_0=0$: there is $t_n\in(-1,0)$ such that
$$e^{-1}-\sum_{k = 0}^{n}  \frac{(-1)^{k}}{k!}=R_n(x)=\frac{e^{t_n}}{(n+1)!}\left(-1\right)^{n+1}.$$
Hence, as $n$ goes to infinity,
$$|I_n|=n!e\cdot \frac{e^{t_n}}{(n+1)!}\leq \frac{e}{(n+1)}\to 0.$$
A: Using integration by parts, we obtain:
$$I(n)=\int_0^1x^ne^xdx=\left[x^ne^x\right]_0^1-n\int_0^1x^{n-1}e^xdx$$
so we could say that:
$$I(n)=\left[x^ne^x\right]_0^1-nI_{n-1}=\left[x^ne^x\right]_0^1-\frac{dI}{dn}$$
You could maybe try to solve this using the differential equation (not sure on this):
$$I+I'=e$$
However if we continue with the pattern of using IBP we notice:
$$I(n)=\left[x^ne^x-nx^{n-1}e^x\right]_0^1+n(n-1)\int_0^1x^{n-2}e^xdx$$
$$=\left[x^ne^x-nx^{n-1}e^x+n(n-1)x^{n-2}e^x\right]_0^1-n(n-1)(n-2)\int_0^1x^{n-3}e^xdx$$
If we were to do this $n$ times we would get:
$$I(n)=\left[e^x\left(\sum_{i=0}^n\frac{(-1)^in!x^{n-i}}{(n-i)!}\right)\right]_0^1-n!\int_0^1e^xdx$$
$$=e\sum_{i=0}^n\frac{(-1)^in!}{(n-i)!}-en!$$
if we notice that $en!$ is just the first term of this series, we can say that:
$$I(n)=e\sum_{i=1}^n\frac{(-1)^in!}{(n-i)!}$$
