How to calculate recursive series? $r_1 = r_0 (1 + e)$

$r_2 = r_1(1 + 2e)$

$r_3 = r_2(1 + 3e)$

$r_n = r_{n-1}(1 + ne)$

Can I get $r_n$ in closed form (a single formula) in terms of $r_0$ and $e$ ? 

Also, as n tens to infinity what will $r_{n}$ converge to ? I guess it will also go to infinity. (given e is smaller than 1).

I have to equate $r_n$ to a term to solve for $e$. 

I can't think of any formula for (1+e)(1+2e)(1+3e)...(1+ne). Is it too difficult to analyse. I guess the higher powers will be minimal as e &lt; 1.

Thanks a lot.
You can divide everything by $r_0$, then multiply it back in at the end, so you are looking for $\prod_{i=1}^n(1+ie)$.  If $e \gt 0$ the infinite sequence does not converge as all the terms are greater than $1+e$.  If $e \lt 0$ the terms in the infinite product become large and negative, so the product gets very large in absolute value and oscillates in sign.  We need to consider finite products only.  

You can collect the terms, saying $$\prod_{i=1}^n(1+ie)=1+\sum_{i=1}S_i(n)e^n$$  $S_i(n)$ is the sum of products of all combinations of $i$ different numbers taken from $\{1,2,3,\dots n\}$  $S_i=\frac 12n(n+1)$, the usual triangular number.  To evaluate $S_2$, imagine adding in the terms where the numbers are the same, in which case we have $S_1^2$, so $S_2(n)=S_1(n)^2-\sum_{i=1}^ni^2=\frac 14n^2(+1)^2-\frac 16n(n+1)(2n+1)$  One can keep going, but it gets messier. They are known as Newton's identities
The limit:

Remark:$e$ seems not equal to 2.718... so we have to look at the case $1+n_0 e =0$ !

if $e$ not in $\{ -\frac1{k}, k \in \mathbb{N}^*\}$, and $r_0 \neq 0$, then 
 $r_n \to \pm\infty$ depending of $e$ and $r_0$.

if $e= -\frac1{k}$, then $r_n =0$ for all $n \ge k$.

if $r_0=0$, $r_n =0$ for all $n \in \mathbb{N}$.

The closer form:

(suppress.Thanks to Did). The coefficients are easily evaluated.

Let $a_{n,k}=0$ for $k \gt n$ and $a_{0,0}=1$ and $r_n = r_0 \sum_{k=0}^n a_{n,k} e^k$, then we have $r_{n+1} = r_0 ((n+1)e+1)\sum_{k=0}^n a_{n,k} e^k$ hence $a_{n+1,k}=a_{n,k-1}(n+1)+ 2*a_{n,k}$.

$a_{n+1,k}=a_{n,k-1}(n+1) +2 a_{n,k}$.