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  "body": "\n## Proof\n\nWe compute the coefficients of the finite partial products, and we show that they stabilise at $p(n)$.\n\n1. All series are [[formal power series|072W67KFYK]] in $\\Z[\\![x]\\!]$. For $k \\ge 1$ let $g_k = \\sum_{m \\ge 0} x^{km}$. So $[x^j] g_k = 1$ if $k \\mid j$, and $[x^j] g_k = 0$ otherwise.\n2. By the [[product formula|0H6KWAYS87]], $[x^n](x^k g_k) = [x^{n-k}] g_k$ for $n \\ge k$, and $0$ for $n \u003c k$. So $x^k g_k = \\sum_{m \\ge 1} x^{km}$.\n3. Hence $(1 - x^k) g_k = g_k - x^k g_k = 1$. So $1 - x^k$ is a unit and $\\frac{1}{1 - x^k} = g_k$.\n4. For $N \\in \\N$ let $P_N = g_1 g_2 \\cdots g_N$, with $P_0 = 1$. The infinite product $\\prod_{k \\ge 1} \\frac{1}{1 - x^k}$ means the series whose coefficient of $x^n$ is $[x^n] P_N$ for all $N \\ge n$.\n5. Steps 6 to 13 show that this is well defined, and compute it. Let $c_N(n)$ be the number of tuples $(m_1, \\dots, m_N) \\in \\N^N$ with $\\sum_{k=1}^{N} k\\, m_k = n$.\n6. Claim: $[x^n] P_N = c_N(n)$ for all $N, n \\in \\N$. We argue by induction on $N$.\n7. For $N = 0$, $P_0 = 1$ has $[x^n] P_0 = 1$ if $n = 0$ and $0$ otherwise. Only the empty tuple has length $0$, and its sum is $0$. So $c_0(n) = [x^n] P_0$.\n8. Let $N \\ge 1$. By the product formula and step 1, $[x^n] P_N = \\sum_{i + j = n} [x^i] P_{N-1} \\cdot [x^j] g_N = \\sum_{m \\ge 0,\\ Nm \\le n} [x^{n - Nm}] P_{N-1}$.\n9. By the induction hypothesis this is $\\sum_{m \\ge 0,\\ Nm \\le n} c_{N-1}(n - Nm)$. Sorting the tuples of length $N$ by their last entry $m_N = m$ shows that this sum is $c_N(n)$.\n10. A [[partition|0404R9P9J3]] of $n$ with all parts at most $N$ is determined by the numbers $m_k$ of its parts equal to $k$, for $1 \\le k \\le N$. These satisfy $\\sum_k k\\, m_k = n$, and each such tuple gives one partition.\n11. So $c_N(n)$ is the number of partitions of $n$ with all parts at most $N$.\n12. Let $N \\ge n$. Each part of a partition of $n$ is at most $n \\le N$. So $c_N(n)$ is the number of all partitions of $n$, which is the [[partition function|001CX45SWQ]] value $p(n)$.\n13. By steps 6 and 12, $[x^n] P_N = p(n)$ for every $N \\ge n$. So the coefficients in step 4 do not depend on $N$, and the infinite product is well defined.\n14. Two series with the same coefficients are equal, by the definition of the [[coefficients|0M98XYVX6S]]. Hence $\\prod_{k \\ge 1} \\frac{1}{1 - x^k} = \\sum_{n \\ge 0} p(n)\\, x^n$, the [[generating function|0HH6JX5WZZ]] of $p$.\n\n## Idea\n\nThe factor $\\frac{1}{1 - x^k} = 1 + x^k + x^{2k} + \\dots$ records how many parts equal $k$: choosing $x^{k m_k}$ means taking $m_k$ parts of size $k$.\nA term of $x^n$ in the expanded product is then exactly a partition of $n$. Factors with $k \u003e n$ only contribute their constant term $1$ to the coefficient of $x^n$.\n",
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