We will prove several congruences modulo a power of a prime such as $$
\sum_{0<k_1<...<k_{n}<p}\leg{p-k_{n}}{3} {(-1)^{k_{n}}\over k_1... k_{n}}\equiv

{lll} -{2^{n+1}+2\over 6^{n+1}} p B_{p-n-1}({1\over 3}) &\pmod{p^2} &{if $n$
is odd} -{2^{n+1}+4\over n6^n} B_{p-n}({1\over 3}) &\pmod{p} &{if $n$ is even}.
$$ where $n$ is a positive integer and $p$ is prime such that $p>\max(n+1,3)$.