<s>[INST] Diophantine equation $1 + \sum_{j=1}^{n-1}\left(j \prod_{k=1}^j x_k\right) = \prod_{j=1}^n x_j$ What are the positive solutions $(x_1,x_2,\ldots,x_n)$ for the Diophantine equation:

$$1 + \sum_{j=1}^{n-1}\left(j \prod_{k=1}^j x_k\right) = \prod_{j=1}^n x_j$$
 [/INST] It seems the following. 
It can be easily checked by induction, that the equation has a simple solution $x_k=k$.
I investigated the equation for small values of $n$. It seems that its solutions can be obtained when we consequently consider admissible by divisibility values for $x_1,\dots,x_{n-1}$, branching our consideration. For instance, by this way we obtain
$n=1$. $1=x_1$ 
So $x_1=1$.
Solutions:


*

*$(1)$


$n=2$. $1+x_1=x_1x_2$. 
So $x_1|1$ and $x_1=1$, $x_2=2$.
Solutions:


*

*$(1,2)$


$n=3$. $1+x_1+2x_1x_2=x_1x_2x_3$. 
So $x_1|1$ and $x_1=1$. Then $2+2x_2=x_2x_3$. 
So $x_2|2$ and $x_2=1$ or  $x_2=2$.
Suppose that $x_2=1$. Then $x_3=4$.
Suppose that $x_2=2$. Then $x_3=3$.
Solutions:


*

*$(1,1,4)$

*$(1,2,3)$


$n=4$. $1+x_1+2x_1x_2+3x_1x_2x_3=x_1x_2x_3x_4$. 
So $x_1|1$ and $x_1=1$. Then $2+2x_2+3x_2x_3=x_2x_3x_4$. 
So $x_2|2$ and $x_2=1$ or  $x_2=2$.
Suppose that $x_2=1$. Then $4+3x_3=x_3x_4$. 
So $x_3|4$ and $x_3=1$ or  $x_3=2$ or $x_3=4$.
Suppose that $x_3=1$. Then $x_4=7$. 
Suppose that $x_3=2$. Then $x_4=5$. 
Suppose that $x_3=4$. Then $x_4=4$. 
Suppose that $x_2=2$. Then $3+3x_3=x_3x_4$. 
So $x_3|3$ and $x_3=1$ or  $x_3=3$.
Suppose that $x_3=1$. Then $x_4=6$. 
Suppose that $x_3=3$. Then $x_4=4$. 
Solutions:


*

*$(1,1,1,7)$ 

*$(1,1,2,5)$

*$(1,1,4,4)$

*$(1,2,1,3)$

*$(1,2,3,4)$


One may write a program which continues this consideration for larger $n$.
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