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  "body": "\n## Proof\n\n1. Write $f = \\sum_{m \\ge 0} a_m X^m$ and $g = \\sum_{m \\ge 0} b_m X^m$. By the definition of the [[coefficient|0M98XYVX6S]], $a_m = [X^m] f$ and $b_m = [X^m] g$.\n2. In $R[\\![X]\\!]$, multiplication is the Cauchy product, by the definition of [[formal power series|072W67KFYK]]. So $f g = \\sum_{n \\ge 0} c_n X^n$ with $c_n = \\sum_{i + j = n} a_i b_j$.\n3. Here the sum runs over the pairs $(i, j) \\in \\N^2$ with $i + j = n$. Hence $[X^n](fg) = c_n = \\sum_{i+j=n} [X^i] f \\cdot [X^j] g$. This is the first equality.\n4. The map $i \\mapsto (i, n - i)$ is a bijection from $\\set{0, 1, \\dots, n}$ to the set of pairs $(i, j) \\in \\N^2$ with $i + j = n$. Its inverse is $(i, j) \\mapsto i$.\n5. Reindexing the finite sum of step 3 along this bijection gives $\\sum_{i+j=n} [X^i] f \\cdot [X^j] g = \\sum_{i=0}^{n} [X^i] f \\cdot [X^{n-i}] g$. This is the second equality.\n\n## Idea\n\nThe formula is the definition of the product, read through the coefficient maps. The second form only lists the pairs $(i, j)$ with $i + j = n$ by their first entry.\n",
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