The [Bell numbers](http://mathworld.wolfram.com/ComplementaryBellNumber.html) $B(n)$ can be given as a sum of the (signed) [Stirling numbers of the second kind](http://mathworld.wolfram.com/ComplementaryBellNumber.html) $S(n,k)$ as $B(n)=\sum\_{k=0}^nS(n,k)$. There are also the so-called [complementary Bell numbers](http://mathworld.wolfram.com/ComplementaryBellNumber.html) defined by
$$B\_1(n):=\sum\_{k=0}^n(-1)^kS(n,k).$$



> 
> **QUESTION.** Given any prime $p$ and a positive integer $k$, is it true that
>  $$B\_1(n+p^k)\equiv B\_1(n+1)-kB\_1(n) \mod p?$$
> 
> 
> 


**EDIT.** In the definition of $B\_1(n)$ there was a typo: $(-1)^n$ was meant to be $(-1)^k$.



Congruence for complementary Bell numbers
The definition of the complementary Bell numbers should be
$$B\_1(n):=\sum\_{k=0}^n(-1)^kS(n,k).$$


Define the polynomials $B\_n(x)$ by 
$$B\_n(x)=\sum\_{k=0}^nx^kS(n,k),$$
so $B\_1(n) = B\_n(-1)$.
These polynomials are called *Bell polynomials* or *exponential polynomials*. Christian Radoux proved the congruence
$$ B\_{n+p^k}(x) \equiv B\_{n+1}(x) + (x^p+x^{p^2}+\cdots+x^{p^k})B\_n(x) \pmod p,$$
where congruence is coefficientwise as polynomials in $x$. (See Anne Gertsch and Alain M. Robert, 
[Some congruences concerning the Bell numbers](http://www.emis.de/journals/BBMS/Bulletin/bul964/Robert-Gertsch.pdf). 
Bull. Belg. Math. Soc. Simon Stevin 3 (1996), no. 4, 467–475, for references and another proof.)
The congruence for the complementary Bell numbers is the case $x=-1$. (Touchard's congruence is the case $x=1$.)


