Factorial Trouble

$10^n$ is actually the number of trailing zeroes of 1000!

$f(1000)= \left\lfloor{\frac{1000}{5}}\right\rfloor + \left\lfloor{\frac{1000}{5^2}}\right\rfloor+ \left\lfloor{\frac{1000}{5^3}}\right\rfloor +\left\lfloor{\frac{1000}{5^4}}\right\rfloor$

$=200+40+8+1=249$

There are 249 trailing zeroes, it means the number is $\displaystyle\large{10^{\boxed{249}}}$

In the expansion of $1000!$ , there are 249 trailing zeroes. So, if there are 249 zeroes are at the back of the number, then it is a must that it is divisible by 10. If $1000!$ is divisible by 10, then the number of trailing zeroes of the number is equal to the value of power in $10^n$ . So, the answer is 249.

The no. of trailing zeroes in 1000! is 249. So, the largest value of n is 249