Be careful with your arithmetic!

Nice Problem!

Here is my solution.

Notice that every $(5k)^{5k}$ contributes $5k$ to the total exponent of $5$ ; every $(5^2k)^{5^2k}$ contribute additional $5^2k$ to the total exponent of $5$ $\ldots{ }\ldots{ }\ldots$ and every $(5^ik)^{5^ik}$ contribute additional $5^ik$ to the total exponent of $5$ , as long as $5^ik \leq 100$ .

The same is for any prime $p$ .

Now we can see,

For any prime $p$ , the highest integer exponent $e$ so that $p^e | \prod_{k=1}^{n} k^k$ = $\sum_{i=1}^{\lfloor \log_p n \rfloor} \left( p^i \sum_{j=1}^{ \lfloor \frac{n}{p^i} \rfloor } j \right)$

This way, I calculated $e=1300$ for $p=5$ . I was intuitively sure that $e\geq1300$ for $p=2$ .

So, number for trailing $0$ s is 1300 .

The product $P = \prod_{k=1}^{100} k^k = 2^{p_2} \times 3^{p_3} \times 5^{p_5} ... 83 \times 89 \times 97$ , where the latter is the product of prime factors and $p_n$ are the power of the prime. Since the number of trailing zeros $n_z$ of $P$ comes from $10^{n_z} = (2 \times 5)^{n_z}$ . This means that $n_z$ is the smaller of $p_2$ and $p_5$ or $n_z = \min (p_2, p_5)$ . It is obvious that $p_2 > p_5$ . Therefore, the number of trailing zeros $n_z = p_5$ . Since every one of $\{5$ , $10$ , $15$ , $20$ , $25$ , $30$ , $35$ , $40$ , $45$ , $50$ , $55$ , $60$ , $65$ , $70$ , $75$ , $80$ , $85$ , $90$ , $95$ , $100 \}$ contributes 1 power of 5 and $\{25$ , $50$ , $75$ , $100 \}$ contributes an additional 1 power of 5, we have:

$\begin{aligned} n_z & = \sum_{k=1}^{20} 5k + \sum_{k=1}^4 25k \\ & = 5 \times \frac {20(21)}2 + 25 \times \frac {4(5)}2 \\ & = 1050 + 250 \\ & = \boxed{1300} \end{aligned}$

The expression can be written as $\large \displaystyle\prod_{r=1}^{100} r^r = 1^1\times 2^2\times\cdots \times 99^{99} \times 100^{100} \quad \small\color{#3D99F6}{\text{Equation 1}}$

Since we require only trailing zeros , we need to find the power of $5$ in $\color{#3D99F6}{\text{ (1) }}$ . $r$ must be a multiple of $5$ .

$\large \displaystyle\prod_{i=1}^{20} (5i)^{5i} = 5^{5+10+15+20+25\times2+\cdots+45+50\times2+\cdots+70+75\times2+\cdots95+100\times 2} \\ \large \implies 5^{5(1+2+3+4\cdots+19+20)+25(1+2+3+4)} \\ \text{Trailing zeroes(Power of 5) : } 1050+250 = \boxed{1300}$

Cleary, the powers of $2$ are much more greater than the powers of $5$ here, so it is sufficient to say that the we are only looking for the maximum power of $5$ that divides the expression.

There are $20$ natural numbers divisible by 5 on $[1,100]$ . So, their sum is $5+10+15+..+100 = 1050$ Next, let us consider those powers of $5$ that repeat (i.e. $25=5^{2} , 75= 5^{2}(3) , 50 = 5^{2}(2) and 100= 5^{2}(4)$ )

Since we already 1 copy each on the latter, we get an answer of $5+10+15+20+...+100 + 25+50+75+100 = 1300$

Every multiple of 5 give same number of zero with itself, and so do the multiple of 25. So, the number of zero is (5+10+15+...+100)+(25+50+75+100)=1300