Intuitive reason why $\sqrt[n]n\to 1$ as $n\to\infty$? We are aware of the limit
$$
\lim_{n\to\infty}\sqrt[n]n = 1;
$$
is there any geometric or otherwise intuitive reason to see why this limit holds?

Edit: I am adding some context, since this question was previously put on-hold, and I think one of the main reasons was that it was poorly motivated. From theorem 8.1 of Baby Rudin, suppose the series
$$
\sum_{n=0}^\infty c_nx^n
$$
converges for $|x|<R$, and define
$$
f(x) = \sum_{n=0}^\infty c_nx^n \qquad (|x|<R). \tag{1}
$$
Among other conclusions, the function $f$ is differentiable in $(-R,R)$, and
$$
f'(x) = \sum_{n=0}^\infty nc_n x^{n-1} \qquad (|x|<R). \tag{2}
$$
Rudin uses the fact that $\sqrt[n]n\to 1$ as $n\to\infty$ to justify that the series in $(1)$ and the series in $(2)$ have the same radius of convergence. I recognized the limit, but it is just such a nice combination of $n$ and the $n$th-root, that I thought there ought to be some nice intuitive way to understand it, hence this question.

If "intuitive reason" contains arguments which are natural, then the following may be of interest.
A recurring argument about convergence of sequences are through subsequences. For example, to see that $\sqrt[n]{a}$ must converge to $1$ if it converges at all (which it does, since it is decreasing) notice that $\sqrt[2n]{a}$ must converge to the same limit $L$, since it is a subsequence. But $(\sqrt[2n]{a})^2=\sqrt[n]{a}$, and therefore $L^2=L \implies L=1$, since it obviously cannot be $0$.
This is just to exemplify a natural argument which commonly applies. Trying to apply this directly in the case $n^{1/n}$ (taking the subsequence $2n$) has problems since the $\sqrt[n]{2}$ will become $1$. Therefore, we must take a subsequence which is increasing faster. It is natural to consider $n^2$. We then have that, if $n^{1/n}$ converges, then $n^{2/n^2}=((n^{1/n})^{1/n})^2$ converges to the same limit. But since $n^{1/n}$ is assumed to be convergent, we have that for sufficiently large $n$:
$$1<((n^{1/n})^{1/n})^2 < ((L+1)^{1/n})^2.$$
By the squeeze theorem, $n^{2/n^2}$ then converges to $1$, and hence so does $n^{1/n}$. In fact, this is a full proof, except that we are supposing that $n^{1/n}$ converges (but it does, since this sequence is eventually decreasing).
PS: This argument has sparkles of the philosophy of the Cauchy condensation test.
