Making $121$ with five $0$s

Question

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?

Answer 1

I'm not sure if this type of solution is what you're looking for, but this sort of problem is pretty trivial if you don't restrict the set of allowed operators somehow:

$$ 121 = \tan \arccos \underbrace{\sin \arctan \sin \arctan \cdots \sin \arctan}_{121^2 \textrm{ copies of}\sin \arctan} \cos 0$$

See USAMO 1995.2.

Answer 2

Note:

If we could define the quadruple factorial formula with some symbols be can get $120$ by using only three $0$'s, because quadruple factorial is formed by formula $\frac{(2n)!}{n!}$.

Reference link here

Answer 3

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. $$$$