Enumeration with factorial

Question

Find all triples (a, b, c) of positive integers such that $a! +b! =c!$ Justify your answer.

How can I justify it, I seem cannot find a solution.

Answer 1

$1!+1!=2!$ is the only solution because

if either $a\ge c$ or $b \ge c$ then $a!+b! >c!$

if $a<c$ and $b<c$

then we can use the fact that $n!>2(n-1)!$ for $n>2$

then for $n>2$ $$ a!+b! \le 2(c-1)! < c! $$