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  "body": "\n## Proof\n\n1. Let $p$ be an odd prime, $x, y$ integers with $p \\nmid x$, $p \\nmid y$, $p \\mid x - y$, and $n \\ge 1$. Here $\\nu_p$ is the [[p-adic valuation|04VCPS30ZE]].\n2. If $x = y$, then $x^n - y^n = 0$. Both sides equal $\\infty$, since $\\nu_p(0) = \\infty$. Assume from now on that $x \\ne y$.\n3. For integers $X, Y$ and $N \\ge 1$, put $S_N(X, Y) = \\sum_{j=0}^{N-1} X^{N-1-j} Y^j$. Expanding the product, the terms cancel in pairs, and $X^N - Y^N = (X - Y) \\, S_N(X, Y)$.\n4. Claim A: if $p \\nmid X$ and $p \\mid X - Y$ and $p \\nmid N$, then $p \\nmid S_N(X, Y)$. Indeed $Y \\equiv X \\pmod p$, so each term is congruent to $X^{N-1}$, by [[powers|0J5VGJYYC3]] and [[products|072D5G4SK6]] of congruences.\n5. So $S_N(X, Y) \\equiv N X^{N-1} \\pmod p$. By [[Euclid's lemma|051VYKA1EX]], $p \\nmid N X^{N-1}$. This proves claim A.\n6. Claim B: if $p \\nmid X$ and $p \\mid X - Y$, then $\\nu_p(S_p(X, Y)) = 1$. Write $Y = X + t$ with $p \\mid t$.\n7. Let $1 \\le j \\le p - 1$. By the [[binomial theorem|0EHWPWGHB2]], $Y^j = X^j + j X^{j-1} t + t^2 w_j$ for an integer $w_j$. Since $p^2 \\mid t^2$, we get $X^{p-1-j} Y^j \\equiv X^{p-1} + j X^{p-2} t \\pmod{p^2}$.\n8. This also holds for $j = 0$, where both sides equal $X^{p-1}$. Summing over $j$ with the [[sum rule for congruences|00FWBN7F0Y]], $S_p(X, Y) \\equiv p X^{p-1} + X^{p-2} t \\sum_{j=0}^{p-1} j \\pmod{p^2}$.\n9. Here $\\sum_{j=0}^{p-1} j = p \\cdot \\frac{p-1}{2}$, and $\\frac{p-1}{2}$ is an integer because $p$ is odd. Since $p \\mid t$, the second term is divisible by $p^2$.\n10. So $S_p(X, Y) \\equiv p X^{p-1} \\pmod{p^2}$. Thus $p \\mid S_p(X, Y)$. As $p \\nmid X^{p-1}$, also $p^2 \\nmid S_p(X, Y)$. This proves claim B.\n11. Let $X, Y$ satisfy $p \\nmid X$, $p \\nmid Y$, $p \\mid X - Y$ and $X \\ne Y$. By claim B, $S_p(X, Y) \\ne 0$, so $X^p \\ne Y^p$. By step 3 and the [[valuation of a product|0850JJMJYZ]], $\\nu_p(X^p - Y^p) = \\nu_p(X - Y) + 1$.\n12. For the same $X, Y$ and $p \\nmid N$, claim A gives $\\nu_p(S_N(X, Y)) = 0$. So $\\nu_p(X^N - Y^N) = \\nu_p(X - Y)$.\n13. For $i \\ge 0$, let $x_i = x^{p^i}$ and $y_i = y^{p^i}$. By Euclid's lemma $p \\nmid x_i$ and $p \\nmid y_i$. By step 3, $x_i - y_i$ is a multiple of $x - y$, so $p \\mid x_i - y_i$.\n14. Claim: $x_i \\ne y_i$ and $\\nu_p(x_i - y_i) = \\nu_p(x - y) + i$. We argue by induction on $i$. The case $i = 0$ is step 2.\n15. If the claim holds for $i$, then step 11 applies to $X = x_i$, $Y = y_i$. Since $x_{i+1} = x_i^p$ and $y_{i+1} = y_i^p$, it gives $x_{i+1} \\ne y_{i+1}$ and $\\nu_p(x_{i+1} - y_{i+1}) = \\nu_p(x - y) + i + 1$.\n16. Write $n = p^a b$ with $a = \\nu_p(n)$ and $p \\nmid b$. Then $x^n - y^n = x_a^b - y_a^b$.\n17. By step 12 with $X = x_a$, $Y = y_a$, $N = b$, and then step 14, $\\nu_p(x^n - y^n) = \\nu_p(x_a - y_a) = \\nu_p(x - y) + \\nu_p(n)$.\n\n## Idea\n\nFactor $x^n - y^n = (x - y)(x^{n-1} + \\dots + y^{n-1})$. When $x \\equiv y \\pmod p$, the second factor is about $n x^{n-1}$ modulo $p$. So it adds no factor of $p$ when $p \\nmid n$, and for $n = p$ it adds exactly one; oddness of $p$ kills the next correction term. Each factor $p$ of $n$ therefore adds one to the valuation.\n",
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