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 
 A: We show there are no solutions when $x\gt 8$. The smaller $x$ can be dealt with by computing.
First note that $z+1$ cannot be even. For if $x\ge 4$, then our sum is congruent to $3$ modulo $5$.
Now show that $z+1$ must be $2$. For if $x=8$ then our sum is divisible by $9$ but by no bigger power of $3$. Any higher factorial is divisible by $3^4$, so for $x\ge 9$, the highest power of $3$ that divides our sum is $3^2$.  
A: 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
A: Hint:
$1!+2!+...x! = y^{z+1}$
$y^{z+1} \ge y^2$
$1!+2!+ \dots +x! \ge y^2$ 
Equality holds iff $x=3$ or $1$, 
Why? When $x \ge 4$, $1!+2! \dots x! \equiv 3 \mod 10$.
Now consider the cases: 
$y^{z+1} = y^{3k}$ or $y^{3k+1}$ or $y^{3k+2}$, for $k$ belonging to Natural numbers.
A: You can easily check Fact $1$: $\displaystyle \sum_{i=1}^{x}i!$ is eqaul to $y^{z+1}$ for $x=3,1$(for $1\le x\le 8).$This has to be checked explicitly by calculating $\displaystyle \sum_{i=1}^{x}i!$ for each $x$.
$\displaystyle \sum_{i=1}^{x}i!=1!+2!+\displaystyle \sum_{i=3}^{x}i! \equiv0$ (mod $3$)(for $x\ge 3$)
So for $x\ge 3,3|y^{z+1}$(if at all the equation holds).$\Rightarrow 3|y$.So let $y=3y_1$
For $x>8$,and considering that the equation holds,
$27|x!$ and $\displaystyle \sum_{i=1}^{8}i!\equiv 9 $ mod $27\dots(*)$.   
At first I will prove that $z+1\ge3$ is not possible,
Let if possible $z+1=3+k,k\ge0$
So we have $y^{z+1}=(3y)^3.(3y)^{k}\dots (1)$
(1) would imply $27|y^{z+1}$ but this contradicts $(*)$
So $(z+1)\ge 3$ is not possible.
So the only possibility is $z+1=2$
Now for $x\ge5$,
$\displaystyle \sum_{i=1}^{x}i!\equiv 3$ mod $10$
but $y^2$ is not equivalent to $3$ mod 10 for all $y\in N$. So $z+1=2$ is also not possible.
So Fact $1$ along with the above implies that the equation is not possible for any other $x$ except $1,3$. Hence we are done.
