If $n=n^{\frac{1}{n}}$, I first assert that for an $n>1$, then $n^{\frac{1}{n}}$ is larger than 1

If $n>1$ then, raising both sides to the power of $1/n$ yields $n^{\frac{1}{n}}>1^{\frac{1}{n}}$. So $n^{\frac{1}{n}}$ is larger than 1 no matter what the value of $1/n$. Therefore, one could make $n^{\frac{1}{n}}=1+R$. As $n^{\frac{1}{n}}$ is always larger than 1, $R$ is always a positive real number. Now, raising both sides to the power $n$ one obtains:

${\displaystyle n=1+\mathrm{nR}+\frac{1}{2}n(n-1)R^2+\dots .}$

(Incidentally, $n>\mathrm{nR}$, therefore $R<1$ and so $n^{\frac{1}{n}}<2$)

As n is an integer larger than one, at least the third term of expansion must exist. It is clear that:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{n}& >\frac{1}{2}n(n-1)R^2\\ \multicolumn{1}{c}{\frac{2n}{n(n-1)}}& >R^2\\ \multicolumn{1}{c}{\sqrt{\frac{2}{n-1}}}& >R\end{array}.}$

So $R\to 0$ as $n\to \infty$ and

${\displaystyle n^{\frac{1}{n}}=1+R<1+\sqrt{\frac{2}{n-1}}.}$

So $n^{\frac{1}{n}}\to 1$ as $n\to \infty$.

As a corollary, I will show that $\sqrt{2/(n-1)}$ grows smaller for successive values of $n>1$. Assume that for some positive $B$, $\sqrt{2/(n-1)}<\sqrt{2/(n+B-1)}$. Now if this is to be true $1/(n-1)<1/(n+b-1)$, therefore $n-1>n+B-1$, therefore $B<0$, contrary to our conditions. Therefore the assertion that for some positive B the value of the function increases is fallacious. This will prove useful later.

For $n=1,\hspace{0.5em}2,\hspace{0.5em}$ etc. we have $n=1,\hspace{0.5em}{2}^{1/2},\hspace{0.5em}{3}^{1/3},\hspace{1em}{4}^{1/4}=2^{1/2}\hspace{0.5em}...$ and we see that the values increase to a maximum of $3^{1/3}$ and then start to decrease.

Now to prove that 3 is the value of $n$ pertaining to the maximum of this discrete function, I will find an $n$ such that $1+\sqrt{2/(n-1)}$ is less that the cube root of 3. From that point onwards, it is known that for successive values $\sqrt{2/(n-1)}$ decreases, therefore beyond this point the value of $n^{\frac{1}{n}}$ decreases; more importantly, is less than $3^{\frac{1}{3}}$. For $n=19$ this holds true, as it will for subsequent values. The ïn-between" values have been checked and are less than $3^{\frac{1}{3}}$. If $n\ge 19$ then

${\displaystyle n^{\frac{1}{n}}<1+\sqrt{\frac{2}{18}}=\frac{4}{3}<3^{\frac{1}{3}}}$