How many 0's are in the end of this expansion? How many $0's$ are in the end of:
$$1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4.... 99^{99}$$
The answer is supposed to be $1100$ but I have absolutely NO clue how to get there.
Any advice?

There as many zeros as the number of the factors of 5 in the product:
So we need to check the subproduct
$$
5^5\cdot 10^{10}\cdots 95^{95}
$$
We have $5+10+15+\cdot+95=5(1+2+\cdots+19)=950$. And we have missed the case where we have two factors: $25,50,75$. So we have another $25+50+75=150$. Altogether $950+150=1100$.
