令 f(x)=∏i=1p−1(x−i)=xp−1+sp−2xp−2+⋯+s1x+s0f(x)=\prod_{i=1}^{p-1}(x-i)=x^{p-1}+s_{p-2}x^{p-2}+\cdots+s_1x+s_0f(x)=∏i=1p−1(x−i)=xp−1+sp−2xp−2+⋯+s1x+s0。由费马定理 xp−1−1≡f(x)(modp)x^{p-1}-1\equiv f(x)\pmod pxp−1−1≡f(x)(modp)。比较系数得到 1≤i≤p−21\le i\le p-21≤i≤p−2 时 p∣sip\mid s_ip∣si，且 s0≡−1(modp)s_0\equiv-1\pmod ps0≡−1(modp)。