Hint: Use integration by parts. EDIT: Try several values of $n$. $$ \int {x e^x dx} = (x - 1)e^x + C $$ $$ \int {x^2 e^x dx} = (x^2 - 2x + 2)e...

You could use the generating function approach. $$ \eqalign{\int_0^X e^{tx} e^x \ dx &= \frac{e^{(1+t)X} - 1}{1+t}\cr &= \sum_{k=0}^\infty (-1)^...

I find it a little difficult for me to guess the solution by trying several $n$. I would like to do it as following: $$\begin{align}\int x^ne^xdx&=...

$ \int{x^ne^xdx}=e^x\sum^n_{k=0}{{\left(-1\right)}^{n-k}\frac{n!}{k!}x^k}+C$ Solution: $ \int{xe^xdx}=\ ?$ $ u=x\to \ \frac{du}{dx}=1\to du=dx$...

A short proof: $\left(1+\frac{x}{n}\right)^n = e^{n\log\left(1+\dfrac{x}{n}\right)}$ Since $\log(1+x) = x + O(x^2)$ when $x \to 0$, we have $n\log(...

$$e^{\ln{(1 + \frac{x}{n})^n} }=e^{n \ln(1+\frac{x}{n})}$$ $$\lim_{n \to +\infty} (1 + \frac{x}{n})^n =\lim_{n \to +\infty} e^{n \ln(1+\frac{x}{n})...

You can use the binomial series expansion. For example: $$\left(1+\frac{x}{n}\right)^n =1+ \frac{n}{1!}\left(\frac{x}{n}\right)^1+\frac{n(n-1)}{2!}...