Recurrence Integration

$\Large \text{ Let } I_n = \int_0^1 e^x(x-1)^ndx$

$\Large \text{ Applying By Parts }$

$\Large I_n = \Big(x-1\Big)^ne^x\bigg|_0^1 -\int_0^1n\Big(x-1\Big)^{n-1}e^xdx$

$\Large \Rightarrow I_n = -(-1)^n - nI_{n-1} \\ \Large\Rightarrow I_n = -nI_{n-1} - (-1)^n$

$\Large \text{ Recursively putting } I_{n-1} \text{ in terms of } I_{n-2} \text{ and } I_{n-2} \text{ in terms of } I_{n-3} \text{ we get } ,$

$\Large I_n = -n\bigg[-(n-1)I_{n-2} - (-1)^{n-1}\bigg] -(-1)^n \\ \Large \hspace{1cm} = n(n-1)I_{n-2} + n(-1)^{n-1} - (-1)^n \\ \Large \hspace{1cm} = n(n-1)\bigg[-(n-2)I_{n-3} - (-1)^{n-2}\bigg] + n(-1)^{n-1} - (-1)^n \\ \Large \hspace{1cm} = -n(n-1)(n-2)I_{n-3} - n(n-1)(-1)^{n-2} + n(-1)^{n-1} - (-1)^n$

$\Large \text{ By repeatedly putting of every } I_j \text{ in terms of } I_{j-1} \text{ after } k_{th} \text{ steps for some integer }k \text{ we will get },$

$\Large I_n = (-1)^k\frac{n!}{(n-k)!}I_{n-k} + \sum\limits_{l=1}^{k} (-1)^l\frac{n!}{(n-l+1)!}(-1)^{n-l+1} \\ \Large \hspace{1cm} = (-1)^k\frac{n!}{(n-k)!}I_{n-k} + (-1)^{n+1}\sum\limits_{l=1}^{k}\frac{n!}{(n-l+1)!} \\$

$\Large\text{Putting } k = n$

$\Large I_n = (-1)^{n}n!I_0 + (-1)^{n+1}\sum\limits_{l=1}^{n} \frac{n!}{(n-l+1)!} \\$

$\Large I_0 = \int_0^1 e^x(x-1)^0dx = e^x\bigg|_0^1 = e-1 \\ \Large I_n = (-1)^{n}n!(e - 1) + (-1)^{n+1}\sum\limits_{l=1}^{n} \frac{n!}{(n-l+1)!} \\ \hspace{1cm} \Large = (-1)^{n+1}\bigg(n! + \sum\limits_{l=1}^{n} \frac{n!}{(n-l+1)!}\bigg) + (-1)^nn!e \\ \\ \Large\text{Since } n \text{ is finite, first term of } I_n \text{ must be rational number and hence does not contains } e \text{ as a factor. So the only term containing } e \text{ is the second term } i.e. (-1)^nn!e \\ \Large\text{ Comapring } I_n \text{ with the given integral } i.e. 16 - 6e \text{ we get } \\ \Large -6 = (-1)^nn! \\ \Large\text{ On solving, we get } \\ \Large \quad n = 3 \\ \Large\text{ Checking this value for the first term of } I_n \text{ we will get } 16 \\ \Large \text{ Hence the answer is } n = \boxed{3}$