Let $f(n,k)$ be the sum of expressions of the form $x_1 \cdot x_2 \cdot \ldots \cdot x_k$, where the sum counts over all solutions of the equation $x_1 + \ldots + x_k = n$ in natural numbers. Find closed-form for the generating function $S_k(z) = \sum_{n=0}^{\infty} f(n,k) x^n$ and for $f(n,k)$.
Of course, if we have $ x_1 \cdot x_2 \cdot \ldots \cdot x_k $, then no $x$ can be zero, so we actually have partitions $ x_1 + \ldots + x_k = n $ into components $ >0 $. From stars and bars, I know that there are $ \binom{n-1}{k-1} $ such partitions. How can I continue?
Notice that, when multiplying polynomials (or power series), the coefficients get multiplied while the exponents get added. This means that if we represent a set of integers as a sum of terms of the form $az^a$, the product of two such sums can be considered as a sum of all terms of the form $a_a a_b z^{a_a+a_b}$, with $a_a$ from the first set and $a_b$ from the second set.
In order to represent that we may use any integers to construct the sums and products we are studying, let's define $I(z) = \sum_{i=0}^{\infty} iz^i = 0z^0 + 1z^1 + 2z^2 + ...$
Let's look at the behaviour of $I(z)^k$:
This power of a power series can be seen as a sum of powers, where the coefficient of the nth power of z is of the form $x_{1,1}x_{1,2}...x_{1,k}+x_{2,1}x_{2,2}...x_{2,k} + ... +x_{\binom{n-1}{k-1},1}x_{\binom{n-1}{k-1},2}...x_{\binom{n-1}{k-1},k}$
so $S_k(z) = I(z)^k$
Now we have to find a closed form for $I(z)$, but this is trivial if we notice that each term of the series is $az^a = z(az^{a-1}) = z \frac{d}{dz} z^a$, so $I(z) = z \frac{d}{dz} (1+z+z^2+...) = z \frac{d}{dz}\frac{1}{1-z} = \frac{z}{(1-z)^2}$
We get that $S_k(z) = \frac{z^k}{(1-z)^{2k}}$
A study of the derivatives of $S_k(z)$ will probably give you an easier way to find $f(n,k)$.