$p^\alpha Cj$ is divisible by $p$

Question

Let $p$ be a prime. For $1\leq j \leq p^\alpha -1$, the binomial coefficient $p^\alpha Cj$ is divisible by $p$.

How?

I know that for $\alpha=1$ the result is true.But how can we generalize it?

Answer 1

Since we know its true for $\alpha=1$, we know that $(a+b)^p=a^p+b^p$ mod $p$. Iterating this, we have $(1+x)^{p^n}=1+x^{p^n}$ mod $p$ as polynomials with coefficients in $\mathbb{Z}/p\mathbb{Z}$. Thus, noting that $(1+x)^{p^n}=\sum_{i=0}^{p^n} (p^nCi) x^i$ as a polynomial, we have $p^nCi=0$ mod $p$ for all $1\leq i\leq p^n-1$.