<s>[INST] 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
 [/INST] 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.
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