Question related to the simplification of a nested radical $\sqrt[n]{\frac{\sqrt[n]{a}}{a^2}\dots}$ 
Given that $a\in \mathbb R^+\setminus\{0\}$ and $n\in \mathbb N$, $n>2$, find the denominator of the irreducible fraction that represents the exponent for $a$ resulting from the simplification of the following nested radical expression:
  $$
\sqrt[\leftroot{+4}\uproot{-35}\scriptstyle n]{
\frac{
\sqrt[\leftroot{+4}\uproot{-25}\scriptstyle n]{
\frac{
\sqrt[\leftroot{+2}\uproot{-15}\scriptstyle n]{
\frac{
\begin{array}{c}
\vdots\\ 
\sqrt[n]{\frac{\sqrt[n]{\frac{\sqrt[n]{a}}{a^2}}}{a^3}}\\
\vdots\\
\end{array}
}{a^{n-2}}
}
}{a^{n-1}}
}
}{a^{n}}
}
$$

This is a question from a math contest. Correct answer unknown. I have a problem at the end in the definition of the denominator of the "irreducible" fraction.
My attempt:
First, by analyzing the intimidating expression, starting from $n=2$ and greater, we can find that it is equivalent to
$$a^{S(n)},~~S(n)=\frac{1-V(n)}{n^n},~~V(n)=2n+3n^2+4n^3+\ldots+n\times n^{n-1}.$$
A simpler expression for $V(n)$ can be found noticing that
$$V(n)-nV(n)=2n+Q(n)-n^{n+1},~~Q(n)=n^2+n^3+\ldots+n^{n-1}.$$
It is easy to see that $Q(n)=\frac{n^2-n^n}{1-n}$ following that
$$V(n)=\frac{2n+\frac{n^2-n^n}{1-n}-n^{n+1}}{1-n}=\frac{(1-n)(2n-n^{n+1})+n^2-n^n}{(1-n)^2}.$$
Finally, we can we get 
$$a^{S(n)},~~S(n)=\frac{1-V(n)}{n^n}=\frac{(1-n)^2-(1-n)(2n-n^{n+1})-n^2+n^n}{n^n(1-n)^2}$$
Another possible way to express the exponent $S(n)$ is 
$$S(n)=\frac{n^{-n}(1-4n+2n^2)+1+n-n^2}{(1-n)^2}$$
Questions: (a) this development is correct up to the definition of $S(n)$? (b) are there ways to simplify it? (c) how to uniquely define the irreducible fraction representing the exponent in this case, so that the denominator is uniquely defined? Note: some numerical exploration suggests that the asked denominator is $n^n$.

Your $S(n)$ is correct. 
For $n=3$, it is easy to see that the asked denominator is $3^3$.
For $n\ge 4$, we have
$$\begin{align}&\frac{1-4n+2n^2+n^n(1+n-n^2)}{(1-n)^2}\\\\&=\frac{-n^n(n^2-2n+1+n-2)+2n^2-4n+2-1}{(n-1)^2}\\\\&=-n^n+2+\frac{-n^{n-1}(n^2-2n+1-1)-1}{(n-1)^2}\\\\&=-n^n+2-n^{n-1}+\frac{n^{n-1}-1}{(n-1)^2}\\\\&=-n^n-n^{n-1}+2+\frac{n^{n-2}+n^{n-3}+\cdots +n+1}{n-1}\\\\&=-n^n-n^{n-1}+n+1+\sum_{j=1}^{n-3}(n-2-j)n^{j}\end{align}$$
So, we get
$$S(n)=\frac{-n^n-n^{n-1}+n+1+\displaystyle\sum_{j=1}^{n-3}(n-2-j)n^{j}}{n^n}$$
It is easy to see that the numerator is coprime to the denominator.
Therefore, the answer is $\color{red}{n^n}$.
