# Prove that $p\mid \binom{p}{k},\ 0< k< p$

Prove that: $$p \,\,\left|\, {p \choose k} \right., \quad 0< k \lt p$$ if $p$ is prime.

how to prove that with direct proof?

• Hint: think of the explicit definition of $\displaystyle\binom{p}{k}$ as a fraction. How many times does $p$ divide the numerator? How many can it divide the denominator? – Steven Stadnicki Nov 28 '12 at 18:40
• can you explain more? – World Nov 28 '12 at 18:45
• Be careful that $k>0$ since $\binom{p}{0}=1$ and then it is not true that $p|1$. – Sebastien B Nov 28 '12 at 18:48
• Also k < p for the same reason. – Gautam Shenoy Nov 28 '12 at 18:51
• @GautamShenoy You are right, I had not seen that both the inequalities were large in the title. – Sebastien B Nov 28 '12 at 19:41

Write out what the binomial coefficient is: $${p\choose k}=\frac{p!}{k!(p-k)!}.$$ $p$ divides the numerator since it has a factor of $p$, but $p$ can't divide the denominator because it is the product of integers smaller than $p$ and $p$ is prime.
This means that $p$ does not appear in the prime factorisation of the denominator, thus you can't simplify the $p$ factor that is on the numerator.
• To complete your argument you could say that $\binom{n}{k}$ is an integer : math.stackexchange.com/questions/11601/… – Sebastien B Nov 28 '12 at 18:55