Proof

Let $(a_n)_{n\in\mathbb N}$ be a real sequence defined by $a_n=\sqrt[n]{a}$ for some positive real number $a > 0$ and all $n\in\mathbb N$.
By definition, the nth-root can be written as the generalized power of $x$, i.e. as $\sqrt[n]{a}=\exp_x\left(\frac 1n\right)$, i.e. the exponential function to the general base $x$.
Since the exponential function of general base is continuous, we have $$\lim_{n\to\infty} a_n=\lim_{n\to\infty}\sqrt[n]{a}=\lim_{n\to\infty}\exp_a\left(\frac 1n\right)=\exp_a\left(\lim_{n\to\infty}\frac 1n\right),\quad\quad( * )$$ which follows from the exchangeability of the limit of function values with the function value of the limit of arguments for continuous functions.
Since it was proven that $\lim_{n\to\infty}\frac 1n=0,$, we have that $( * )$ equals $\exp_a(0).$
By definition, $\exp_a(0)=\exp(0\cdot \log (a))=\exp(0).$
Since it was proven that $\exp(0)=1$, it follows that $\lim_{n\to\infty}\sqrt[n]{a}=1.$
∎

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Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983