Paneer Definite Masala 11

As @展豪 張 told,

$\begin{aligned}I_n &= e^x(x-1)^n|_0^1- n\int\cdots\\ &= e^x\left((x-1)^n - n(x-1)^{n-1}+\cdots +(-1)^n n!)\right)_0^1+ \end{aligned}$

Since at the limit $1$ , All the initial terms in $e^x ()$ cancel, we get $(-1)^n n!e = -6e$

Hence $n = 3$

Let $\displaystyle I_n=\int_0^1 e^x(x-1)^n\mathrm{d}x$
Using integration by parts, $\displaystyle I_n=\int_0^1 e^x(x-1)^n\mathrm{d}x=e^x(x-1)^n \vert_0^1-n\int_0^1e^x(x-1)^{n-1}\mathrm{d}x=(-1)^{n+1}-I_{n-1}$
$I_0=\int_0^1 e^x \mathrm{d}x=e^x \vert_0^1=e-1$
$I_1=(1)-(1)(e-1)=2-e$
$I_2=(-1)-(2)(2-e)=-5+2e$
$I_3=(1)-(3)(-5+2e)=16-6e$ as desired.
$\therefore n=3$

I will give a rigorous approach although this one can be solved by hit and trial after a few steps.
Let $I_n=\displaystyle \int_0^{1} e^x(x-1)^ndx$ $I_n=(x-1)^ne^x \left. \right|_0^1 -n \displaystyle \int_0^1 e^x (x-1)^n$ $=(-1)^{n+1}-n \cdot I_{n-1}$
Now, define another sequence $J_n=I_n+ \lambda \text{ (you will see why)}$ $J_n+n \cdot J_{n-1} + (n+1)\lambda=(-1)^{n+1}$ $J_n=-nJ_{n-1}+(-1)^{n+1}-(n+1)\lambda$ We are free to choose our $\lambda$ , so let it be $\lambda=\dfrac {(-1)^{n+1}}{n+1}$ $\implies \dfrac {J_n}{J_{n-1}}=-n \implies J_n=J_0 (n!) (-1)^{n} \text{ (Use telescoping products)}$
Substitute $J_n$ for $I_n$ , simplify, compare both sides for coefficient of $e$ to get $(-1)^n (n!) =-6 \implies n=3.$