TITLE: Making $121$ with five $0$s
QUESTION [19 upvotes]: So I say this puzzle online a few days ago and found it quite interesting. The original question was 

Make $120$ using only five $0$s. 

Well, I said to myself, this is utterly trivial. Note that $$ 120 = 5! = (0! + 0! + 0! + 0! + 0!)!. $$
But what if we want to do it for an arbitrary number $n$ and an arbitrary number of $0$s, $m$. That is: we want to make $n$ using only $m$ zeroes. Clearly, using my solution above, we can make $n=m!$ using $m$ zeroes. 
For $n=121$ and $m=5$, this is tougher and I can't seem to find a solution. Does anybody want to try to take on some general cases?

REPLY [3 votes]: A Perelman-like solution might suffice :
$$121 = -\log_2 \log_2 \underbrace{\sqrt{\sqrt{\sqrt{\cdots\sqrt{(0!+0!)!}}}}}_\text{121 copies square roots}$$
Which uses only two $0$'s, next best to betaveros. $$$$