Does Lucas' Theorem hold for $p$-adic integers? More specifically, does it specifically hold for the case that, given a $p$-adic integer: $$x = x_0 + x_1p + x_2p^2 + \cdots = \sum_{i=0}^\infty x_i p^i$$ where each $x_i \in \{0,1,\ldots,p-1\}$, is it true that for each $n$ a natural number or zero, $$\binom{x}{p^n} \equiv x_n \bmod p\ ?$$ If it is true, does that mean that it holds if we do not put the integer in canonical form?
For example, suppose $a$ is a p-adic integer with a multiplicative inverse and that the canonical representations for $a$, $-a$, and $1/a$, are given by,
$$a = \sum_{i=0}^\infty ( a_i p^i ) \ \land \ -a = \sum_{i=0}^\infty ( b_i p^i ) \ \land \ 1/a = \sum_{i=0}^\infty ( c_i p^i )$$
that is, for each $i$, $a_i , b_i , c_i \in \{0,\ldots,p-1\}$. Then is it true that
$$ \binom{-a}{p^n} = (-1)^{p^n} \binom{a+p^n-1}{p^n} \equiv b_n \mod p $$
$$ \binom{1/a}{p^n} = \frac{\prod_{i=0}^{p^n-1} (1-a \times i)}{a^{p^n} \times (p^n)!} \equiv c_n \mod p $$
For the case that $p=2$, I've checked the first few digits for the negation operation $-a$ and the first two or three digits for $1/a$ and it seems to hold. But is it true in general?
Clarification concerning interpretation of binomial coefficients: For powers of p in the denominator of binomial coefficients and fractions modulo p, common p factors of the numerator and denominator are to be canceled out before interpreting the fraction mod p so that division by zero will not occur. If common p factors are canceled out and there are still factors of p in the denominator then the fraction is undefined mod p.
Any help or information is very much appreciated. Thank you for your time and have a great day.