Find $\lim_{n\to\infty} n\int_1^e x^a(\log_ex)^ndx$

Substituting $x=e^t$ and integrating by parts give $$ n\int_1^e {x^a \log ^n xdx} = n\int_0^1 {e^{(a + 1)t} t^n dt} \\ =(n+1)\int_0^1 {e^{(a + 1)t} t^n dt}- \int_0^1 {e^{(a + 1)t} t^n dt} \\ = e^{a + 1} - (a + 1)\int_0^1 {e^{(a + 1)t} t^{n + 1} dt} - \int_0^1 {e^{(a + 1)t} t^n dt} . $$ Now $$ \left| {(a + 1)\int_0^1 {e^{(a + 1)t} t^{n + 1} dt} } \right| \le \left| {a + 1} \right|\max(1,e^{a + 1}) \int_0^1 {t^{n + 1} dt} \\ = \left| {a + 1} \right|\max(1,e^{a + 1}) \frac{1}{{n + 2}} $$ and $$ 0 < \int_0^1 {e^{(a + 1)t} t^n dt} \le \max(1,e^{a + 1}) \int_0^1 {t^n dt} = \max(1,e^{a + 1}) \frac{1}{n + 1}. $$ Consequently, the limit in question is $e^{a+1}$.

I thought it might be instructive to present an approach that appeals to the Dominated Convergence Theorem. To that end we proceed.

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Let $f_n(a)$ be given by

$$f_n(a)=n\int_1^e x^a\log^n(x)\,dx\tag1$$

Enforcing the substitution $x\mapsto e^{1-x/n}$ in $(1)$ reveals

$$\begin{align} f_n(a)&=e^{a+1}\int_0^\infty \xi_{[0,n]}(x)\,\,e^{-(a+1)x/n}\left(1-\frac xn\right)^n\,dx\tag1 \end{align}$$

where $\xi_{[0,n]}(x)$ is the indicator function.

Inasmuch as $\left|\xi_{[0,n]}(x)\,\,e^{-(a+1)x/n}\left(1-\frac xn\right)^n\right|\le \max(1,e^{-(a+1)})e^{-x}$ and $\int_0^\infty \max(1,e^{-(a+1)})e^{-x}\,dx<\infty$, the Dominated Convergence Theorem guarantees that

$$\begin{align} \lim_{n\to\infty}f_n(a)&=e^{a+1}\int_0^\infty \lim_{n\to \infty}\left(\xi_{[0,n]}(x)\,\,e^{-(a+1)x/n}\left(1-\frac xn\right)^n\right)\,dx\\\\ &=e^{a+1}\int_0^\infty e^{-x}\,dx\\\\ &=e^{a+1} \end{align}$$

And we are done!