Time for Factorials

WLOG $a_n\geq a_{n-1} \geq \dots \geq a_1$ . Note that $X>a_n$ since factorials is a monotonicaly increasing function. We have $\sum_{i=1}^{n} a_i ! \leq n(a_n !)$ $\frac{\sum_{i=1}^{n} a_i !}{a_n !} \leq n$ $\frac{\sum_{i=1}^{n} a_i !}{a_n !} \in\ \{2,3,4.\dots,n\}$ We note that for some $X!\neq n$ cannot work since $\lambda!\geq 1$ . Thus, the possible solution is $X=n,a_n=n-1$ which yields $n-1=a_n= a_{n-1} = \dots = a_1$