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Why $$\binom{mp-1}{p-1} \equiv 1 \pmod {p^3}?$$

I tried by induction on $m$:

$\bullet$ $m=0$ $$\binom{-1}{p-1} = (-1)^{p-1}=1 \equiv 1 \pmod {p^3}$$

$\bullet$ $m=1$ $$\binom{p-1}{p-1} =1 \equiv 1 \pmod {p^3}$$

$\bullet$ induction hylpothesis: $$\binom{mp-1}{p-1} \equiv 1 \pmod {p^3}$$

$\bullet$ $m+1$ $$\binom{(m+1)p-1}{p-1}= \binom{(m+1)p}{p}-\binom{(m+1)p-1}{p}$$ but I am not able to move forward. Have you any hints?

Thank you so much

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    $\begingroup$ Assuming $p$ is supposed to be prime, then I try $p=2, m=2$ and get $\binom 31=3\neq 1 \bmod 8$ $\endgroup$ – Mark Bennet Jan 27 at 16:54
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    $\begingroup$ The $p \ge 5$, $m=2$ cases are Wolstenholme's theorem. (The primes $p=2$ and $p=3$ already don't work.) Experimentally, though, the result does seem true for all $p \ge 5$ and all $m$. $\endgroup$ – Misha Lavrov Jan 27 at 17:17
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The necessary tools can be found in Glaisher's 1900 paper "Congruences relating to the sums of products of the first $n$ numbers and to other sums of $n$ products", which is more or less available here. (It is the first article in the issue.) Glaisher only proves the $m=2$ case this way, I think - but the method works for all cases.

Define the polynomial $$ f(x) = (x+1)(x+2)\dotsb (x+p-1) = \left[{p\atop 1}\right] + \left[{p\atop 2}\right]x + \dots + \left[{p\atop p}\right]x^{p-1} $$ where $\left[{p\atop k}\right]$ denotes the unsigned Stirling number of the first kind. (Glaisher does not use this terminology or notation.)

The idea is that $\binom{mp-1}{p-1}$ can be written as the ratio $\frac{f((m-1)p)}{f(0)}$, and so it is only necessary to show that $$ f((m-1)p) \equiv f(0) \pmod{p^3} $$ and also that neither is divisible by $p$, to prove your theorem. (Well, $f(0)$ is just $(p-1)! = \left[{p\atop 1}\right]$, so we know all about it, and it's not divisible by $p$.) This goes in $3$ steps:

  1. We have $$f((m-1)p) \equiv \left[{p\atop 1}\right] + \left[{p\atop 2}\right]((m-1)p) + \left[{p\atop 3}\right]((m-1)p)^2 \pmod {p^3},$$ since all other terms have a factor of $p^3$ in them.
  2. Actually, when $p>2$, $\left[{p\atop 3}\right]$ (as well as all other coefficients except $\left[{p\atop 1}\right]$) is divisible by $p$, so the quadratic term also vanishes, and we get $$f((m-1)p) \equiv \left[{p\atop 1}\right] + \left[{p\atop 2}\right]((m-1)p) \pmod {p^3}.$$ To see the divisibility by $p$, Glaisher compares the coefficients in $f(x+1)$ and $(x+1)f(x)$, and uses the known divisibility properties of binomial coefficients.
  3. Actually, when $p>3$, $\left[{p\atop 2}\right]$ is divisible by $p^2$, so the linear term vanishes, and we get $$f((m-1)p) \equiv \left[{p\atop 1}\right] \pmod{p^3}.$$ To see this, Glaisher expands $f(-p)=\left[{p\atop 1}\right]$, and then all terms except the $\left[{p\atop 2}\right]$ term have a factor of $p^2$ in them.

This completes the proof, since we've concluded that $f((m-1)p) \equiv f(0) \pmod{p^3}$, and since neither is divisible by $p$, we can divide and get $\frac{f((m-1)p)}{f(0)} \equiv 1 \pmod{p^3}$.

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