HMMT-Nov 2017 guts, Problem 5

Define a sequence $\left\{a_{n}\right\}$ by $a_{1}=1$ and $a_{n}=\left(a_{n-1}\right) !+1$ for every $n>1$. Find the least $n$ for which $a_{n}>10^{10}$.

Solution: We have $a_{2}=2, a_{3}=3, a_{4}=7, a_{5}=7 !+1=5041$, and $a_{6}=5041 !+1$. But \[ 5041 !+1 \gg 5041 \cdot 5040 \cdot 5039>10^{10} \] Hence, the answer is 6 . $\fbox{6}$.

Answer: 6