socrates wrote: Determine the least $n\in\mathbb{N}$ such that $n!=1\cdot 2\cdot 3\cdots (n-1)\cdot n$ has at least $2010$ positive factors. $14!$ has $2*36^2$ divisors and $13!$ has $1584$ [ stupid by guessing but found in a few minutes]

More generally if the prime factorisation of $N$ is $N=p_1^{a_1}*p_2^{a_2}*...*p_n^{a_n}$ for distinct primes $p_1,p_2,...,p_n$ then the number of factors of N is $(a_1+1)(a_2+1)...(a_n+1)$. Now $13!=2^{10}*3^{5}*5^{2}*7*11*13$ so has $11*6*3*2*2*2=1584$ factors while $14!=2^{11}*3^{5}*5^{2}*7^{2}*11*13$ so has $12*6*3*3*2*2=2592$ factors just as SCP said.

It's an increasing sequence so answer is $14$