We have $$\sum\limits_{0\leqslant a_1<a_2<\cdots<a_k<n}\binom{n}{a_1}\prod\limits_{p=2}^{k}\binom{n-a_{p-1}}{a_p-a_{p-1}}=k!{n+1\brace k+1}$$ where instead of $\{a_1,a_2,\cdots,a_k\}$ we place all possible combinations of $\{0,1,2,\cdots,n-1\}$ without repetitions, for ex. $n=4, k=3$, so all possible combinations are $$\{0,1,2\},\{0,1,3\}, \{0,2,3\}, \{1,2,3\}$$ then $$\binom{4}{0}\binom{4}{1}\binom{3}{1}+\binom{4}{0}\binom{4}{1}\binom{3}{2}+\binom{4}{0}\binom{4}{2}\binom{2}{1}+\binom{4}{1}\binom{3}{1}\binom{2}{1}=60=3!{5\brace 4}$$
How can we prove it?
Well, $${n+1\brace k+1}k!={n\brace k}k!+{n\brace k+1}(k+1)k!={n\brace k}k!+{n\brace k+1}(k+1)!,$$ Recall that $k!{n\brace k}$ counts the number of surjective functions from $[k]$ to $[n]$ The left hand side is basically saying choose $a_1$ from $n,$ so that $1$ is send to them then choose $a_2-a_1$ from $n-a_1$ that $2$ will be send to. Etc. At the end of the process there is two options, either you finish selecting all of them or they had left. If they were left, then those are gonna be the image of $k+1.$