Evaluate $\lim_{n\to \infty}n\int_2^e{(\ln x)^n}dx$

By integrating by parts ($u'(x) = \frac{(\ln x)^{n}}{x}$, $v(x) = x$) we have $$ \int (\ln x)^n dx = \frac{(\ln x)^{n+1} x}{n+1} - \frac{1}{n+1} \int (\ln x)^{n+1} dx$$ $$ n\int_2^e (\ln x)^n dx = \frac{n\big(e- 2(\ln 2)^{n+1})\big)}{n+1} - \frac{n}{n+1} \int_2^e (\ln x)^{n+1} dx$$ $$ \lim_{n\rightarrow\infty} n\int_2^e (\ln x)^n dx = e - \lim_{n\rightarrow\infty} \int_2^e (\ln x)^{n+1} dx$$ and you have already proven that the second term is $0$.