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Let $B(n)$ denote the Bell's number which is the number of the equivalence relation which can be defined on a set of cardinality $n$.

While I was trying to solve a problem, I reached another result;

$$B(p^k)\equiv k+1 \ mod \ p$$

Is this result evident or trivial ? Any comments and remark are welcome.

Note: My field is not Number theory, that is why I am not famialiar with the tools and result in number theory. If this question is something trivial, please excuse me.

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  • $\begingroup$ It is well-known and there are more general results for arbitrary polynomials of a prime variable $p$ instead of $p^k$. The rhs depends on a polynomial only, if I remember correctly. $\endgroup$
    – Andrew
    Commented Jan 5, 2015 at 20:08
  • $\begingroup$ @Andrew: Can you please give some reference ? $\endgroup$
    – mesel
    Commented Jan 5, 2015 at 20:10

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This follows from the well-known Touchard's congruence (here for a random reference). Following your notation, the congruence is:

$$B(n+p^k)\equiv kB(n)+B(n+1) \mod \ p$$

Taking $n=0$:

$$B(p^k)\equiv kB(0)+B(1) \mod \ p$$

And since $B(0)=B(1)=1$,

$$B(p^k)\equiv k+1 \mod \ p$$

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