Binomial Coefficient involving Power of Prime/Proof 1

Theorem

$\dbinom {p^n k} {p^n} \equiv k \pmod p$

where $\dbinom {p^n k} {p^n}$ is a binomial coefficient.

Proof

From Prime Power of Sum Modulo Prime we have:

$(1): \quad \paren {a + b}^{p^n} \equiv \paren {a^{p^n} + b^{p^n} } \pmod p$

We can write this:

$\paren {a + b}^{p^n k} = \paren {\paren {a + b}^{p^n} }^k$

By $(1)$ and Congruence of Powers, we therefore have:

$\paren {a + b}^{p^n k} \equiv \paren {a^{p^n} + b^{p^n} }^k \pmod p$

The coefficient $\dbinom {p^n k} {p^n}$ is the binomial coefficient of $b^{p^n}$ in $\paren {a + b}^{p^n k} = \paren {\paren {a + b}^{p^n} }^k$.

Expanding $\paren {a^{p^n} + b^{p^n} }^k$ using the Binomial Theorem, we find that the coefficient of $b^{p^n}$, the second term, is $\dbinom k 1 = k$.

So:

$\dbinom {p^n k} {p^n} \equiv k \pmod p$

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 2.6$. Algebra of congruences: Example $42 \ (5)$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $11$: The Sylow Theorems: Lemma $11.2$