A reduction formula

1. With $u=x^n$, $v'=e^x$:
$$I_n=\bigl[x^ne^x\bigr]_0^1-n\int_0^1x^{n-1}e^x\,dx=e-nI_{n-1}.$$

2. $I_0=\int_0^1e^x\,dx=e-1$. Then
- $I_1=e-I_0=1$,
- $I_2=e-2I_1=e-2$,
- $I_3=e-3I_2=e-3e+6=6-2e\approx0.5634$.

3. On $[0,1]$, $0<e^x\le e$, so $0<I_n\le e\int_0^1x^n\,dx=\frac e{n+1}\to0$.

Remark. The recurrence $I_n=e-nI_{n-1}$ is *numerically unstable* when run forward: any small error is multiplied by $n$ at each step. To compute $I_{20}$ one should run it backward, from a good estimate of $I_{30}$ for instance.