The $p\mid m$ part of $S(p^h,a)$ is $p^{k-1}S(p^{h-k},a)$ for $h\ge k$ · Prove2Me

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The $p\mid m$ part of $S(p^h,a)$ is $p^{k-1}S(p^{h-k},a)$ for $h\ge k$ · Prove2Me