Combinatorial congruences modulo prime powers

Let p be any prime, and let a and n be nonnegative integers. Let $r\in Z$ and $f(x)\in Z[x]$. We establish the congruence $$p^{\mathrm{deg}f}\sum _{k=r(mod{p}^a)}\left(\genfrac{}{}{0px}{}{n}{k}\right)(-1{)}^{k}f((k-r)\mathrm{/}{p}^a)=0(mod{p}^{\sum _{i=a}^{\mathrm{\infty }}[n\mathrm{/}{p}^i]})$$ (motivated by a conjecture arising from algebraic topology), and obtain the following vast generalization of Lucas’ theorem: If a is greater than one, and $l,s,t$ are nonnegative integers with $s,t<p$, then $$\frac{1}{[n\mathrm{/}{p}^{a-1}]!}\sum _{k=r(mod{p}^a)}\left(\genfrac{}{}{0px}{}{pn+s}{pk+t}\right)(-1{)}^{pk}((k-r)\mathrm{/}{p}^{a-1}{)}^l=\frac{1}{[n\mathrm{/}{p}^{a-1}]!}\sum _{k=r(mod{p}^a)}\left(\genfrac{}{}{0px}{}{n}{k}\right)\left(\genfrac{}{}{0px}{}{s}{t}\right)(-1{)}^k((k-r)\mathrm{/}{p}^{a-1}{)}^l(modp)\mathrm{.}$$ We also present an application of the first congruence to Bernoulli polynomials, and apply the second congruence to show that a p-adic order bound given by the authors in a previous paper can be attained when p=2.