Generalizing $\sum \limits_{n=1}^{\infty }n^{2}/x^{n}$ to $\sum \limits_{n=1}^{\infty }n^{p}/x^{n}$
The general closed form is
$$\displaystyle \sum_{k=1}^{\infty} k^n x^k = \frac{1}{(1 - x)^{n+1}} \left( \sum_{m=0}^{n} A(n, m) x^{m+1} \right)$$
where $A(n, m)$ are the Eulerian numbers. When I have time I will edit with a few more details. If you only want the answer for a particular small value of $n$ then see Section 3 of my notes on generating functions. I will also mention that for a particular value of $n$ one can deduce the answer by using the identity
$$\displaystyle \sum_{k=0}^{\infty} {k+n \choose n} x^k = \frac{1}{(1 - x)^{n+1}}$$
and writing $k^n$ as a linear combination of the polynomials ${k+r \choose r}$ (in $k$), for example using a finite difference table.
Here is a different look:
The differentiating and multiplying by $x$ gives rise to Stirling Numbers of the Second Kind.
Say you denote the operator of differentiating and multiplying by $x$ as $D_{x}$
Then we have that
$$(D_{x})^{n}f(x) = \sum_{k=1}^{n} s(n,k) f^{(k)}(x) x^{k}$$
where $s(n,k)$ is the stirling number of the second kind and $f^{(k)}(x)$ is the $k^{th}$ derivative of $f(x)$.
This can easily be proven using the identity $$s(n,k) = s(n-1,k-1) + k \cdot s(n-1,k)$$
Here is a table for the Stirling numbers of the second kind (from the wiki page):
n/k 0 1 2 3 4 5 6 7 8 9 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1
So in your case, we can start with $ f(x) = \frac{1}{1-x}$ and obtain that
$$ \sum_{k=0}^{\infty} k^{n} x^k = \sum_{r=1}^{n} r! \cdot s(n,r) \frac{x^{r}}{(1-x)^{r+1}}$$
For example, in your case for $n=3$ we get
$$\sum_{k=0}^{\infty} k^3 x^k = \frac{1! \cdot 1 \cdot x}{(1-x)^2} + \frac{2! \cdot 3 \cdot x^2}{(1-x)^3} + \frac{3! \cdot 1 \cdot x^3}{(1-x)^4}$$
$$ = \frac{x(1-x)^2 + 6x^{2}(1-x) + 6x^3}{(1-x)^4} $$
$$ = \frac{x^3 + 4x^2 + x}{(1-x)^4} $$