Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$.

Actually you are done. You already have: $$nI_n=\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}}$$ hence $$\begin{align}\lim_{n\to\infty} nI_n &= \lim_{n\to\infty}\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}} \\ &= \int_0^1 \lim_{n\to\infty} \frac{dt}{\sqrt{1+t^{-2/n}}} \\ &= \int_0^1\frac{dt}{\sqrt{1+1}} = \frac{1}{\sqrt{2}} \end{align}$$

You can exchange limit and integration due to dominated convergence theorem.

You can still use Sandwich theorem, if you haven't yet seen dominated convergence theorem :

Let $ n $ be a positive integer.

As you said, using the substitution $ \left\lbrace\begin{aligned}y&=x^{n}\\ \mathrm{d}y &=n x^{n-1}\,\mathrm{d}x\end{aligned}\right. $, we get : $$\int_{0}^{1}{\frac{n x^{n}}{\sqrt{1+x^{2}}}\,\mathrm{d}x}=\int_{0}^{1}{\frac{y^{\frac{1}{n}}}{\sqrt{1+y^{\frac{2}{n}}}}\,\mathrm{d}y}$$

Meaning, we have : \begin{aligned}\left|\frac{1}{\sqrt{2}}-n I_{n}\right|&=\left|\int_{0}^{1}{\left(\frac{1}{\sqrt{2}}-\frac{y^{\frac{1}{n}}}{\sqrt{1+y^{\frac{2}{n}}}}\right)\mathrm{d}y}\right| \\ &=\left|\int_{0}^{1}{\frac{\sqrt{1+y^{\frac{2}{n}}}-\sqrt{2}y^{\frac{1}{n}}}{\sqrt{2\left(1+y^{\frac{2}{n}}\right)}}\,\mathrm{d}y}\right|\\ &=\int_{0}^{1}{\frac{1-y^{\frac{2}{n}}}{\sqrt{2\left(1+y^{\frac{2}{n}}\right)}\left(\sqrt{1+y^{\frac{2}{n}}}+\sqrt{2}y^{\frac{1}{n}}\right)}\,\mathrm{d}y}\end{aligned}

Since $ \left(\forall y\in\left[0,1\right]\right),\ \sqrt{2\left(1+y^{\frac{2}{n}}\right)}\left(\sqrt{1+y^{\frac{2}{n}}}+\sqrt{2}y^{\frac{1}{n}}\right)\geq \sqrt{2}+\sqrt{2}y^{\frac{2}{n}}\geq 1 $, we have : $ \int\limits_{0}^{1}{\frac{1-y^{\frac{2}{n}}}{\sqrt{2\left(1+y^{\frac{2}{n}}\right)}\left(\sqrt{1+y^{\frac{2}{n}}}+\sqrt{2}y^{\frac{1}{n}}\right)}\,\mathrm{d}y}\leq\int\limits_{0}^{1}{\left(1-y^{\frac{2}{n}}\right)\mathrm{d}y} $, and thus : $$ \left|\frac{1}{\sqrt{2}}-n I_{n}\right|\leq\int_{0}^{1}{\left(1-y^{\frac{2}{n}}\right)\mathrm{d}y}=\frac{2}{n+2}\underset{n\to +\infty}{\longrightarrow}0 $$

Hence : $$ \lim_{n\to +\infty}{nI_{n}}=\frac{1}{\sqrt{2}} $$