Integer Partition (Restricted number of parts)

Note that

$$x^k\prod_{i=1}^k\frac1{1-x^i}=\left(\prod_{i=1}^{k-1}\frac1{1-x^i}\right)\cdot\frac{x^k}{1-x^k}\tag{1}$$

is the product of the following series:

$$\begin{align*} &1+x+x^2+x^3+x^4+\ldots+x^n+\ldots\\ &1+x^2+x^4+x^6+x^8+\ldots+x^{2n}+\ldots\\ &1+x^3+x^6+x^9+x^{12}+\ldots+x^{3n}+\ldots\\ &1+x^4+x^8+x^{12}+x^{16}+\ldots+x^{4n}+\ldots\\ &\qquad\qquad\qquad\vdots\\ &1+x^{k-1}+x^{2(k-1)}+x^{3(k-1)}+x^{4(k-1)}+\ldots+x^{(k-1)n}+\ldots\\ &x^k+x^{2k}+x^{3k}+x^{4x}+x^{5k}+\ldots+x^{kn}+\ldots \end{align*}$$

Each term of the product will therefore have the form

$$x^{n_1}x^{2n_2}x^{3n_3}\ldots x^{kn_k}=x^{n_1+2n_2+3n_3+\ldots+kn_k}\;,$$

where $n_1,n_2,\ldots,n_{k-1}\ge 0$, and $n_k\ge 1$. This term is $x^n$ if and only if

$$n_1+2n_2+3n_3+\ldots+kn_k=n\;,\tag{2}$$

in which case it corresponds to a partition of $n$ into $n_1$ parts of size $1$, $n_2$ parts of size $2$, and so on. Since $n_k\ge 1$, this partition must have a largest part of size $k$. Conversely, each partition of $n$ with largest part $k$ yields a solution to $(2)$ with $n_k\ge 1$. Thus, the coefficient of $x^n$ in $(1)$ is precisely the number of partitions of $n$ with largest part $k$.

Now look at the Ferrers diagram of any partition of $n$ with largest part $k$. Flip that diagram over its main diagonal, so that the top row becomes the first column and vice versa. You now have the Ferrers diagram of a partition of $n$ with exactly $k$ parts; this partition is the conjugate of the original partition. Every partition of $n$ into exactly $k$ parts arises in this way from a partition of $n$ with largest part $k$, so there are exactly as many partitions of $n$ into exactly $k$ parts as there are with largest part $k$: the former are the conjugates of the latter (and vice versa). Thus, the coefficient of $x^n$ in $(1)$ is also the number of partitions of $n$ into exactly $k$ parts.