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

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 :
$$\frac{1}{2(p-1)}$$
