If $a$, $b$, $c$, $d$, $e$ are positive integers such that $abcde=a+b+c+d+e$, then what is the maximum value of $e$? 
If $a$, $b$, $c$, $d$, $e$ are positive integers such that $abcde=a+b+c+d+e$, then what is the maximum value of $e$?
The answer is $5$.

What I tried:
Let $a\leq b \leq c \leq d \leq e$. Then 
$$abcde \leq 5e \quad\Rightarrow\quad abcd \leq 5$$
For the maximum value of $e$, here $a=b=c=1$ and $d=5$.
So $$e=\frac{a+b+c+d}{abcd-1}=\frac{8}{4}=2$$
But this is not the answer.
How do I solve it? Help me, please.

It's impossible that $a=b=c=d=1.$
Thus, $abcd\geq2$ and
$$e=\frac{a+b+c+d}{abcd-1}\leq5.$$
The equality occurs for $(a,b,c,d,e)=(1,1,1,2,5).$
For a proof of the last inequality we can use the following reasoning.
Let $a=1+x$, $b=1+y$, $c=1+z$ and $d=2+t$.
Thus, $x$, $y$, $z$ and $t$ are non-negatives and we need to prove that:
$$5(x+1)(y+1)(z+1)(t+2)\geq5+x+y+z+t+5,$$which is obvious.
