Let the real sequence $\left(x_{n}\right)_{n\geq1},$defined by $x_{1}>1$ and $x_{n+1}=x_{n}^{n+1}$+$x_{n}+1~$whenever $n\geq1.$Prove that there is a real number $a\neq0$,such that $\displaystyle\lim_{n\rightarrow\infty}$$\frac{x_{n}}{a^{n!}}=1$
That's what I've tried. I don't know if there is a simplier method to solve the exercise. Let $a_n:=x_n^{1/n!}$. We have $x_{n+1}\ge x_n^{n+1}$ hence $a_{n+1}\ge a_n$, increasing sequence. Also we have $x_n\ge 1$ hence $x_n+1\le 2x_n^2\le 2x_n^{n+1}$ hence $x_n^{n+1}+x_n+1\le 3x_n^{n+1}$. This implies $a_{n+1}\le 3^{1/(n+1)!}a_n$. Therefore $a_{n+1}\le a_1 \Pi_1^n 3^{1/(k+1)!}$ and $a_n$ converges to a limit $a>0$. And i suppose that this is the $a$ that we must find.
Please help!!!