# Function whose $n$-th derivative at $x=0$ is $n^3$ / evaluating the power series $\sum_{n=1}^\infty n^3\frac{x^n}{n!}$

I have troubles proving that the power series

$\sum_{n=1}^\infty n^3\frac{x^n}{n!}$

represents the function

$f(x)=e^x(x^3+3x^2+x)$.

My idea was using the identity theorem for power series and the definition of the Taylor series of a function to somehow show that the only function whose $n$-th derivative at $x=0$ equals $n^3$ is the aforementioned function, but right now I can't find the right arguments and I need some hints to point me in the right direction.

Hint: $$\frac{n^3}{n!}=\frac{n^2}{(n-1)!}=\frac{1}{(n-1)!}+\frac{n+1}{(n-2)!}=\frac{1}{(n-1)!}+\frac{3}{(n-2)!}+\frac{1}{(n-3)!}$$
$\textbf{Using generating function method :}$
Define : $$G[x,\lambda]=\sum_{n=0}^{\infty}e^{i\lambda n}_{} \frac{x^{n}_{}}{n!}=e^{x e^{i\lambda}}_{}.$$ Now note $$\sum_{n=1}^{\infty}n^{3}_{}\frac{x^{n}_{}}{n!}=\sum_{n=0}^{\infty}n^{3}_{}\frac{x^{n}_{}}{n!}=\frac{\partial^{3}_{}}{\partial(i\lambda)^{3}_{}}G[x,\lambda]\Big|_{\lambda=0}^{}.$$