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.

Here the issue is that $n \rightarrow \infty$, but for any fixed $x > 0$ we have $x^{1/n} \rightarrow 1$.  So to consider $n^{1/n}$ you have to ask which "wins": the $n$ at the base or the $1/n$ in the exponent.   To think about this, it might help to compare to, say, $(2^n)^{1/n}$.  This tends to (and is equal to) $2$.  The exponent of $1/n$ has the power to take a huge number like $2^n$ and reduce it to a constant.  Since $n$ is much, much smaller than $2^n$, you might expect then that the power of $1/n$ would "win" in the end, giving a result of $1$.  This is not a proof, but it gives the right intuition.
