How prove this limits $\lim_{n\to\infty}\frac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)}{n^2}=\frac{1}{8}$ Interesting Question:

Let denote by $v_{p}(a)$ the exponent of the prime number $p$ in the prime factorization of $a$,
show that
  $$\lim_{n\to\infty}\dfrac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)}{n^2}=\dfrac{1}{8}$$

My some idea: since
$$1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n=\dfrac{(n!)^n}{1!\cdot 2!\cdot 3!\cdots (n-1)!}$$
and it is well know
$$v_{5}(n!)=\lfloor \dfrac{n}{5}\rfloor+\lfloor\dfrac{n}{5^2}\rfloor+\lfloor\dfrac{n}{5^3}\rfloor+\cdots+\lfloor\dfrac{n}{5^k}\rfloor+\cdots=\sum_{i=1}^{\infty}\lfloor\dfrac{n}{5^k}\rfloor$$
so
$$v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)=n\sum_{i=1}^{\infty}\lfloor\dfrac{n}{5^k}\rfloor-\sum_{i=1}^{n-1}\sum_{k=1}^{\infty}\left(\lfloor\dfrac{i}{5^k}\rfloor\right)$$
then I can't it.
Thank you
 A: Exploiting:
$$\sum_{k=1}^{+\infty}\left\lfloor\frac{n}{5^k}\right\rfloor = \frac{n}{4}+O(1)$$
there is left very little to do.
A: I think it is better to count directly. 
We have $5+10+15+...+5[n/5]=5\frac{[n/5]([n/5]+1)}{2}$ counts one factor $5$ from each multiple of $5$ raised to the power the number has.
Then $25+50+75+...+25[n/25]=25\frac{[n/25]([n/25]+1)}{2}$ counts one factor $5$ from each multiple of $25$ raised to the power it appears. Notice the other factor $5$ it provides was already been counted in the previous sum. 
And so on, the next count is $125+250+...+125[n/125]=125\frac{[n/125]([n/125]+1)}{2}$.
We sum now all these sums
$$5\frac{[n/5]([n/5]+1)}{2}+25\frac{[n/25]([n/25]+1)}{2}+125\frac{[n/125]([n/125]+1)}{2}+...$$
It is enough to compute the limit for $n$ a power of $5$, $n=5^N$. The sum becomes
$$5\frac{5^{N-1}(5^{N-1}+1)}{2}+25\frac{5^{N-2}(5^{N-2}+1)}{2}+...+5^N\frac{1(1+1)}{2}\\=\frac{(5^{2N-1}+5^N)+(5^{2N-2}+5^{N})+...+(5^N+5^N)}{2}\\=\frac{5^N\frac{5^N-1}{5-1}+N5^N}{2}\\=\frac{5^{2N}-5^N+4N5^N}{8}$$
We must divide this by $5^{2N}$ and take limit, which is $1/8$. For $n$ moving along powers of 5 plus a bit the work is similar.
A: This seemed like a fun problem and wanted to prove it for general prime p. I use Legendre's formula directly which is,
$$v_p(n!) = \frac{n-s_p(n)}{p-1}$$
$s_p(n)$ being the sum of digits of $n$ when written in base $p$.
$$\frac{1}{n^2}v_p(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)=\frac{1}{n^2}v_p\left(\dfrac{(n!)^n}{1!\cdot 2!\cdot 3!\cdots (n-1)!}\right)$$
This splits up by additivity to be,
$$\frac{1}{n}v_p(n!) - \frac{1}{n^2}\sum_{k=1}^{n-1} v_p(k!)$$
Now using Legendre's formula 
$$\frac{1}{n}\frac{n-s_p(n)}{p-1} - \frac{1}{n^2}\sum_{k=1}^{n-1} \frac{k-s_p(k)}{p-1}$$
A little more algebra and replacing the sum over k with $\frac{n(n-1)}{2}$,
$$\frac{1}{p-1} \left( \frac{1}{2} + \frac{1}{2n}- \frac{s_p(n)}{n} + \frac{1}{n^2}\sum_{k=1}^{n-1} s_p(k)\right)$$
We can put a very cheap upper bound on $s_p(n)$ since the number of digits it has is $\lceil\log_p(n)\rceil$ and the largest a digit can be is $p-1$, so 
$$s_p(n) \le \lceil\log_p(n)\rceil(p-1)$$
This higher bound still has a limit of 0 for the terms containing $s_p$, and thus squeezes out :
$$\frac{1}{2(p-1)}$$
