Non-recursive way to present $ p_{0}=0$, $p_{n+1}=(e+1)p_{n}+e$ for some $e>0 \in \mathbb{R}$.

Question

Is there a non-recursive way to present this function:

$ p_{0}=0$

$p_{n+1}=(e+1)p_{n}+e$ for some $e>0 \in \mathbb{R}$.

Or at least some estimation from the top would satisfy me.

Answer 1

Note that $p_{n+1}+1=(\mathrm e+1)(p_n+1)$ for every $n\geqslant0$ and that $p_0+1=1$ hence $p_n=(\mathrm e+1)^n-1$ for every $n\geqslant0$. Thus, for example, $p_n\lt(\mathrm e+1)^n$ for every $n\geqslant0$.