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  "body": "\n## Proof\n\nWe multiply the sum by $x$ and subtract.\nWe compute in the field $\\mathbb{R}$, see [[the reals form a field|0FGENPMPJX]].\nLet $x \\neq 1$ be real, let $n$ be a natural number, and let $s = \\sum_{k=0}^{n-1} x^k$.\n\n1. If $n = 0$, the sum is empty, so $s = 0$. Also $\\frac{x^0 - 1}{x - 1} = \\frac{0}{x - 1} = 0$. So the formula holds.\n2. Now let $n \\ge 1$. By distributivity, $x s = \\sum_{k=0}^{n-1} x^{k+1} = \\sum_{k=1}^{n} x^k$.\n3. We write $s = 1 + \\sum_{k=1}^{n-1} x^k$ and $x s = \\sum_{k=1}^{n-1} x^k + x^n$.\n4. Subtracting the first equation of step 3 from the second, the common sum $\\sum_{k=1}^{n-1} x^k$ cancels. So $x s - s = x^n - 1$.\n5. By distributivity, $x s - s = (x - 1) s$. So $(x - 1) s = x^n - 1$.\n6. Since $x \\neq 1$, the number $x - 1$ is not $0$, so it has an inverse in the field $\\mathbb{R}$.\n7. Multiplying step 5 by $(x - 1)^{-1}$ gives $s = \\frac{x^n - 1}{x - 1}$.\n\n## Idea\n\nMultiplying by $x$ shifts every power up by one.\nThe shifted sum and the original sum share all terms except $1$ and $x^n$, so their difference is $x^n - 1$.\n",
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