How prove this $\binom{n}{m}\equiv 0\pmod p$ let $p$ is prime number,and such $p\mid n,p\nmid m,n\ge m$
show that
$$p\>\Big|\>\binom{n}{m}$$
I know that: if $p$ is prime number,then
$$\binom{n}{p}\equiv \left[\dfrac{n}{p}\right] \pmod p$$
But I can't prove my problem,Thank you 
 A: We have
$$
{n\choose m}=\frac{n}{m}{n-1\choose m-1}.
$$
Now ${n-1\choose m-1}$ is an integer, and $p$ divides the numerator of $\frac{n}{m}$ but not the denominator. It follows that $p$ divides the numerator of ${n\choose m}$, but since ${n\choose m}$ is an integer this means $p|{n\choose m}$.
A: Let us write $n$ and $m$ in base $p$:
$$n = n_kp^k + n_{k-1}p^{k-1} + \cdots +n_1p+ n_0$$
$$m = m_\ell p^\ell + m_{\ell-1}p^{\ell-1} + \cdots + m_1p+ m_0$$
The fact that $p\mid n$ suggests that $n_0 = 0$ and the fact that $p\nmid m$ suggests $m_0 > 0$. Therefore $$\binom{n_0}{m_0} \equiv 0 \pmod p$$ and the result follows as a direct consequence of Lucas' Theorem.
A: The power of $p$ in $n!$ is
$$a=\lfloor \frac{n}{p} \rfloor+\lfloor \frac{n}{p^2} \rfloor+....$$
The power of $p$ in $m!(n-m)!$ is
$$b=\lfloor \frac{m}{p} \rfloor+\lfloor \frac{n-m}{p} \rfloor+\lfloor \frac{m}{p^2} \rfloor+\lfloor \frac{n-m}{p} \rfloor+....$$
Now, for each $k$ we have
$$\lfloor \frac{m}{p^k} \rfloor+\lfloor \frac{n-m}{p^k} \rfloor \leq \frac{m}{p^k} + \frac{n-m}{p^k} =\frac{n}{p^k}$$
which implies
$$ \lfloor \frac{m}{p^k} \rfloor+\lfloor \frac{n-m}{p^k} \rfloor \leq \lfloor \frac{n}{p^k} \rfloor$$
We now show that for $k=1$ the inequality is strict. Indeed, as $p \nmid m$ we have
$$\lfloor \frac{m}{p} \rfloor <  \frac{m}{p} $$
hence
$$ \lfloor \frac{m}{p} \rfloor+\lfloor \frac{n-m}{p} \rfloor <   \frac{m}{p} + \frac{n-m}{p} =\frac{n}{p}=\lfloor \frac{n}{p} \rfloor $$
This shows that
$$b < a$$
Thus $a-b \geq 1$ and $p^{a-b}| \binom{n}{m}$.
