Proof

By hypothesis, $\sum_{n=0}^\infty a_n$ is an infinite series, and for a fixed¹ positive number $0 < q < 1$ there is an index $N$ such that for all $n\ge N$ the n-th root of absolute values is smaller than or equal $q$, i.e. $\sqrt[n]{|a_n|}\le q$ for all $n\ge N.$
This is equivalent to $|a_n|\le q^n$ for all $n\ge N.$
Since the geometric series $\sum_{n=N}^\infty q_n$ is convergent, it follows from the direct comparison test for absolutely convergent series that $\sum_{n=0}^\infty a_n$ is absolutely convergent.
Now, assume there is an index $N\in\mathbb N$ such that $\sqrt[n]{|a_n|}\ge 1$ for all $n\ge N,$ or at least $\sqrt[n]{|a_n|}\ge 1$ for infinitely many $n\in\mathbb N.$
This is equivalent to $|a_n|\ge 1.$
Therefore, the sequence $(a_n)_{n\in\mathbb N}$ cannot converge to zero.
By contraposition to convergent series implies convergent sequence, the series $\sum_{n=0}^\infty a_n$ is divergent.

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Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition

Please note that you have to find a fixed number $q$ that is $< 1.$ In order to be able to apply the root test, it does not suffice to show that $\sqrt[n]{|a_n|} < 1$ for all $n > N,$ the inequality must be $\sqrt[n]{|a_n|} < q < 1.$ ↩