Find $\lim\limits_{n \to \infty} n \int_2^e (\ln x)^n \, dx$.

As you said $\lim\limits_{n\to \infty}\ln^n 2=0$. Let:

$$I_n = \int_2^e\ln^n x\,dx$$

$I_n$ is decreasing because:

$$I_{n}-I_{n+1}=\int_2^e\ln^n x(1-\ln x)\,dx \geq 0$$

Integrating by parts:

$$ \begin{aligned} \displaystyle\int_2^e \ln^n x\, dx &= \displaystyle\int_2^e \ln^n x \cdot (x)'\, dx\\ &=x\ln^n x\bigg|_2^e -n\int_2^e\ln^{n-1}x\,dx\\ &= e-2\ln^n 2-nI_{n-1} \end{aligned} $$

Therefore $I_n+nI_{n-1}=e-2\ln^n2$. Now, since $I_n$ is decreasing:

$$I_n+nI_{n-1}\geq I_n+nI_n\Rightarrow I_n\leq \frac{1}{n+1}(e-2\ln^n 2)$$

and

$$I_{n+1}+(n+1)I_n\leq I_n+(n+1)I_n\Rightarrow I_n\geq \frac{1}{n+2}(e-2\ln^{n+1}2)$$

Combining the two:

$$\frac{1}{n+2}(e-2\ln^{n+1}2) \leq I_n\leq \frac{1}{n+1}(e-2\ln^n 2)$$

or

$$\frac{n}{n+2}(e-2\ln^{n+1}2) \leq nI_n\leq \frac{n}{n+1}(e-2\ln^n 2)$$

and squeezing $\lim\limits_{n\to \infty}nI_n=e$.

Use $t = \log^{n+1}(x) \implies \frac{1}{n+1}e^{t^{\frac{1}{n+1}}}dt = \log^n(x)dx$:

$$I = \lim_{n\to\infty} \frac{n}{n+1}\int_{\log^{n+1}(2)}^1 e^{t^{\frac{1}{n+1}}}\:dt \longrightarrow \int_0^1 e\:dt = e$$

by dominated convergence.