Prove that $x_n \rightarrow a \Rightarrow (x_n)^{\frac{1}{\alpha}} \rightarrow a^{\frac{1}{\alpha}} $ when $\alpha \in \mathbb{N}$ [duplicate]

One of my friends asked me this question while doing a proof for convergence and this has confused me quite a bit. While this can be proved when $\frac{1}{\alpha} \in \mathbb{Z}$ easily by algebra of limits (specifically, by the result: $\{a_n\}\rightarrow a$ and $\{b_n\}\rightarrow b \Rightarrow \{a_nb_n\} \rightarrow ab$), I am not sure how to go about proving this result when $\frac{1}{\alpha} \in (0,1)$

One approach that I could come up with was by using the fact that the function$ f(x) = x^{\frac{1}{\alpha}}\ $ is continuous everywhere in its domain $\mathbb{R}^{+}$ and hence for every sequence $\{x_n\} \rightarrow a \; , \{f(x_n)\}\rightarrow f(a),$ therefore if $\{x_n\}\rightarrow a$ then $ \{f(x_n)\} = \{x_n^{\frac{1}{\alpha}}\}\rightarrow f(a) = a^\frac{1}{\alpha} $.

However, I and my friend too, are not comfortable using the concept of continuity to show the result and are looking for purely "sequences" based proof.

Also, my intuition tells me that the Bernoulli inequality might come in handy but I can't really pin it down.

Thank you!