I read an article about integer partition which is posted on Wikipedia, and I found out that the generating function of "partitions of n into exactly k parts" can be represented as:
$$\sum_{n\ge0} p_k(n)x^n=x^k\prod\limits_{i=1}^k \dfrac{1}{1-x^i}$$ I tried to understand why generating function can be represented like that but I couldn't. So please help me understand it.
On
By conjugation of the partition (which is the same as reflecting the Ferrers diagram across its main diagonal) one easily sees that there is a bijection between partitions of $n$ into exactly $k$ parts and partitions of $n$ with largest part $k$. The generating function you have written can readily be seen to count the latter (and hence the former, which is what you are after).
Expanding the factors in $x^k \prod_{i=1}^k \frac{1}{1-x^i}$ (as geometric series) you see that the product becomes
$$ x^k (1+x^{1 \cdot 1}+x^{1 \cdot 2}+x^{1 \cdot 3}+ \cdots ) (1+ x^{2\cdot 1} + x^{2 \cdot 2} + x^{2 \cdot 3} + \cdots ) \cdots (1 + x^{k \cdot 1} + x^{k \cdot 2} + x^{k \cdot 3} + \cdots ). $$
From the first bracket you choose the multiplicity of $1$ in your partition, from the second bracket you choose the multiplicity of $2$ and so on up to the multiplicity of $k$. You can not choose multiplicity $0$ for $k$ since the largest part has to be $k$, hence why we multiply by $x^k$ to ensure its multiplicity is at least one.
On
It helps if you look at the product going from $k$ down to $1$.
Then we have at least one part of size $k$ (from the $x^k$), and maybe some more from $\dfrac1{1-x^k}$.
Then $\dfrac1{1-x^{k-1}}$ supplies parts of size $k-1$, and so on until $\dfrac1{1-x}$ provides parts of size $1$.
Then instead of reading the corresponding Ferrers diagram from left to right, we read it from bottom to top, and we see we have exactly $k$ 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.