Given $f(x) = x^n e^{-x}$, show that $\int_0^1 f(x)\, dx$ is equal to a given expression. Consider the function:
$$f : \mathbb{R} \rightarrow \mathbb{R} \hspace{2cm} f(x) = x^n e^{-x}$$
I have to show the following:
$$\int_0^1 f(x) dx = n! \bigg [ 1 - \dfrac{1}{e} \bigg ( 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + ... + \dfrac{1}{n!} \bigg ) \bigg ]$$
I used the notation:
$$I_n = \int_0^1 f(x)dx$$
And by integrating by parts I got the recurrence formula:
$$I_n = - \dfrac{1}{e} + n \cdot I_{n - 1}$$
But I don't have any idea as to how I could show what is asked.

Clearly, the answer is true for $n=0$. 
Now supposing it is true for $n-1$, let us prove for $n$. 
We know thaf $I_n=-\frac{1}{e}+n\cdot I_{n-1}$. 
Expand that out to see that $$I_n=-n!\frac{1}{n!e}+n!(1-\frac{1}{e}(1+...+\frac{1}{(n-1)!}))$$.
Take the common terms, and we see the statement holds for $n$ as well, and we have completed our proof. 
