TITLE: How find this equation $n!+(n+1)!+\cdots+(n+m)!=a^b$ all solutions
QUESTION [7 upvotes]: Find all natural numbers $m,n$ and positive integer $a,b>1$, such
$$n!+(n+1)!+\cdots+(n+m)!=a^b$$
I find  

(1):$$1!+2!+3!=1+2+6=9=3^2$$
  then
  $$n=1,m=2,a=3,b=2$$

Now @Xoff have found other two solution

(2):$$2!+3!=8=2^3$$
  so
  $$n=1,m=1,a=2,b=3$$
(3):$$2!+3!+4!=32=2^5$$
  so
  $$n=2,m=2,a=2,b=5$$

and Now  Peter Košinár Found the fourth solution

(4):$$4!+5!=12^2$$
  so
  $$n=4,m=1,a=12,b=2$$

so I only find this,I think this have other,so How can find all solution,Thank you
This problem is my student ask me,so　I don't kown come from
Thank you
@Ivan Loh  and so on can see this nice problem 
Oh,my student tell me this problem is  from china BBS:http://tieba.baidu.com/p/2760536313
Today I have solve follow problem:
if $x,y,z\in Z$,such
$$1!+2!+3!+\cdots+x!=y^z$$
have only four case
(1):$$x=1,y=\pm 1,z=2n$$
(2):$$x=3,y=\pm 3,z=2$$
(3):$$x\in Z,z=1,y=1+2!+\cdots+x!$$
(4):$$x=1,y=1,z\ge 3,z=2n+1$$

REPLY [4 votes]: I'm not entirely sure why one would ask such a question, but I suspect it might be pretty hard to solve it completely (or maybe not!). On the other hand, it is not difficult to show, for example, that there is only the one known solution with, say, $n > m^2$, by noting that $n!$ divides the left-hand-side and that each prime between $n/2$ and $n$ divides $n!$ exactly once. The product of such primes is roughly of size $e^{n/2}$ which (under the assumption that $n > m^2$) is greater than 
$$
\left( n! + (n+1)! + \cdots + (n+m)! \right)/n!.
$$
This ensures that there is some prime between $n/2$ and $n$ which divides $a^b$ exactly once, contradicting $b > 1$. There are some details required to make this argument explicit, but none of them are too unpleasant.