Proving that $ L = \{ 0^{{2n}\choose{n}} : n\in\mathbb{N} \}$ is not regular
A unary language (i.e., a subset of $0^*$) $L$ is regular iff the set $S = \{ n : 0^n \in L \}$ is eventually periodic. In particular, if $L$ is regular then either $S$ is finite or $S$ has positive density (in the following strong sense: $|S \cap [n]|/n$ converges to a positive number). However, in your case the set $S$ is infinite yet has density zero.

In the same way, we get that the following languages are not regular:
$$
\{0^{\lfloor n\log n \rfloor} : n \in \mathbb N\},
\{0^{n^2} : n \in \mathbb N\},
\{0^{2^n} : n \in \mathbb N\},
\{0^{n!} : n \in \mathbb N\}
$$
and so on.