TITLE: When n! = some Fibonacci number
QUESTION [13 upvotes]: I'm trying to work out following problem:
Let $m$ be some Fibonacci number and $n$ some integer. Find all non-trivial pairs $(n,m)$, where $n! = m$. Trivial pair is pair $(1,1)$.
I know that for every number $a$ there exists some Fibonacci number that's divisible by $a$. I also tried to bruteforce my way trough, but I didn't succeed.
Here is my sub-optimal code:
a = 1
b = 1
mylist = [here are factorials up to 50!]
for i in range(0,100):
c = a+b
a = b
b = c
print(b)
if b in mylist:
    print("found")  

I tried many other things including googling it, but I didn't find anything that I can even start with.
I have highschool level of math knowledge.

REPLY [21 votes]: There are only three terms of the Fibonacci sequence which are equal to a factorial:
$$F_{1}= F_{2} = 0!=1! \quad \mbox{and} \quad F_{3}=2!$$
What is more, F. Luca proved in this paper (published in 1999) that the only nontrivial solutions of the diophantine equation $$F_{n} = m_{1}! \cdots m_{t}!$$
are $F_{3}=2!$, $F_{6}=(2!)^{3}$, and $F_{12} = (2!)^{2}(3!)^{2} = 3! 4!$.

REPLY [14 votes]: On Fibonacci numbers that are factorials, Surajit Rajagopal and Martin Griffiths (2014)
This is behind a paywall, but thanks to SJR the paper can be accessed here. The "simple proof" that for any $n\geq 4$ the Fibonacci number $F_n$ is not a factorial starts on page 14.