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?

To find out the number of zeroes at the end of that huge number, we just have to find out the exponent of $10$ in it.
Since $10=2\cdot 5$, we can just find out the exponent of $5$ in that number as the exponent of $2$ will be more than $5$.
All the numbers which have $5$ as their factor will be counted.
These numbers would be
$$5^5,10^{10},15^{15},\ldots,95^{95}$$
We know that all numbers except $25,50$ and $75$ will only contribute themselves as the exponent, whereas the aforementioned numbers, will contribute double of themselves.
So, we have
$$(5+10+\cdots+95)+(75+50+25)=\frac{19}{2}(5+95) + 150=1100$$     
