Generating function of number of partitions of $n$ into parts at most $k$

Question

I am trying to grasp the intuition behind this example.

Show that $\sum_{n \geq 0} p_{\leq k}(n)x^n = \prod_{i=1}^k \frac{1}{1-x^i}$ where $p_{\leq k}(n)$ denotes the number of partitions of the integer $n$ into parts of size at most $k$.

The book A Walk Through Combinatorics starts off the solution by determining the coefficients of $x^n$ on the right hand side, but I am not really sure how this helps our case. Are we supposed to show that the coefficients of $x^n$ on the right hand side shows the integer partition of $n$ with parts at most $k$?

If so, my question is:

What should one show to prove that the coefficients of $x^n$ demonstrates an integer partition of $n$ with parts no greater than $k$? What do we need to show that we can finally say "coefficients indeed demonstrate an integer partition of $n$ with parts at most $k$?"

Answer 1

We can write the product as

\begin{align*} \prod_{i=1}^k \frac{1}{1-x^i}&=\left(1+x+x^2+\cdots\right)\left(1+x^2+x^4+\cdots\right)\\ &\qquad\cdot \left(1+x+x^2+\cdots\right)\left(1+x^2+x^4+\cdots\right)\\ &\qquad\cdots\\ &\qquad\cdot \left(1+x^k+x^{2k}+\cdots\right) \end{align*} where each factor $\left(1+x^j+x^{2j}+\cdots\right)$ with $1\leq j\leq k$ represents zero, one or more occurrences of a part $j$. We observe the coefficient of $x^n$ in the product provides the number of occurrences to select zero, one or more parts of size $1$ to $k$ which sum up to $n$.

We conclude \begin{align*} \color{blue}{\sum_{n \geq 0} p_{\leq k}(n)x^n = \prod_{i=1}^k \frac{1}{1-x^i}} \end{align*}

Example: We look at an example with small values $n=5$ and $k=4$. Denoting with $[x^n]$ the coefficient of $x^n$ of a series we get \begin{align*} \color{blue}{[x^5]\prod_{i=1}^4 \frac{1}{1-x^i}} &=[x^5]\left(1+x+x^2+\cdots\right)\left(1+x^2+x^4+\cdots\right)\\ &\qquad\qquad\cdot \left(1+x^3+x^{6}+\cdots\right)\left(1+x^4+x^{8}+\cdots\right)\\ &=[x^5]\left(1+x+x^2+x^3+x^4+x^5\right)\left(1+x^2+x^4\right)\left(1+x^3\right)\left(1+x^4\right)\\ &=[x^5]\left(x^5\cdot1\cdot1\cdot1+x^3x^2\cdot1\cdot1+x^2\cdot1\cdot x^3\cdot1\right.\\ &\qquad\qquad\left.+x^1x^4\cdot1\cdot1 +1\cdot1\cdot x^2x^3\cdot1+x^1\cdot 1\cdot 1\cdot x^4\right)\\ &\,\,\color{blue}{=6} \end{align*} corresponding to the $6$ representations of $5$ with parts $1\leq k\leq 4$: \begin{align*} \color{blue}{5}&\color{blue}{=1+1+1+1+1}\\ &\color{blue}{=1+1+1+2}\\ &\color{blue}{=1+1+3}\\ &\color{blue}{=1+2+2}\\ &\color{blue}{=1+1+3}\\ &\color{blue}{=1+4} \end{align*}