how to solve $1!+2!+3!+...+x!=y^{z+1}$where $x,y,z\in \mathbb N$? how to solve the  following equation where $x,y,z\in \mathbb N$ $$1!+2!+3!+...+x!=y^{z+1}$$
Thanks in advance 

must be $x\gt1$ because $x=1,y=1,z+1$ is airbitrary is one solution. we have $2$ case:
case$I$ : $z=1, x\ge 2$ we have $1!+...+x!=y^2$ in mod10 
we have for $x=2$, rightward the statement is $3$ but $ y^2\equiv 0,1,4,5,6,9 \pmod {10}$ so that is no solution. 
if $x=3$ then we have $y=3,z=1$ is solution.
if $x\ge4$ rightward the statement is $3$ but $ y^2\equiv 0,1,4,5,6,9 \pmod {10}$ so that is no solution.(because $10k=n!\equiv 0 \pmod{10}$ for $n \ge 5$)
case$II$: $z\ge 2$ then $3$ divide rightward of the statement because$(1!+2!)(=3), n \ge3 ,n!=3k$ so $3|y^{z+1}$ thus $3|y$ by Euclid lemma for $3$ that is prime.
$y=3k , 3^{z+1}k^{z+1}=27s=y^{z+1} , z\ge2$ for $n \ge9 , n! \equiv 0 \pmod {27}$ so that is enough to show for $x=1,...,8$ that is no solution i.e rightward of the statement $\not \equiv 0 \pmod {27}$ it's easy : we have:
$x=2 \Rightarrow 3\not \equiv 0 \pmod {27}$    ---,$x=3 \Rightarrow 9\not \equiv 0 \pmod {27}$,---$x=4\Rightarrow 6\not \equiv 0 \pmod {27}$,
$x=5 \Rightarrow 18\not \equiv 0 \pmod {27}$ ,---$x=6 \Rightarrow 9\not \equiv 0 \pmod {27}$,---$x=7\Rightarrow 5\not \equiv 0 \pmod {27}$, $x=8 \Rightarrow 9\not \equiv 0 \pmod {27}$,
only solution is $x=3,y=3,z=1$ if $x \gt 1$ and $x \in \Bbb N$ : $$x=3,y=3,z=1,---x=1,y=1,z+1$$ is airbitrary
