This follows from the well-known Touchard's congruence (here for a random reference). Following your notation, the congruence is:
$$B(n+p^k)\equiv kB(n)+B(n+1) \ mod \ p$$$$B(n+p^k)\equiv kB(n)+B(n+1) \mod \ p$$
Taking $n=0$:
$$B(p^k)\equiv kB(0)+B(1) \ mod \ p$$$$B(p^k)\equiv kB(0)+B(1) \mod \ p$$
And since $B(0)=B(1)=1$,
$$B(p^k)\equiv k+1 \ mod \ p$$$$B(p^k)\equiv k+1 \mod \ p$$
