Proving prime $p$ divides $\binom{p}{k}$ for $k\in\{1,\ldots,p-1\}$ Prove if $p$ is a prime then $p \mid \binom pk$ for $k\in\{1,\ldots,p-1\}$
I don't really know where to begin with this one. 
I can see that I have to use the fact that $p$ is prime somewhere - the same is not true for composite numbers, for example $4\nmid 6=\binom42$.
I have checked that this is true for the first few primes: 


*

*$3$ divides $\binom 31=\binom32=3$

*$5$ divides both $\binom 51=\binom 54=5$ and $\binom 52=\binom 53=10$

*$7$ divides $\binom71=\binom76=7$, $\binom72=\binom75=21$ and $\binom73=\binom74=35$.

 A: let $a = (p-1)!$ and $b = k!(p-k)!$ then $\binom pk = p*(a/b)$ so $b*\binom pk = p*a$.
So therefore $p\mid b*\binom pk$ and $p$ does not divide $b$ since it is a product of natural numbers each one being less than $p$. so therefore since $p$ is prime $p\mid \binom pk$.  
A: With a tiny bit of group theory, you can do without using factorisation into primes.
Let $X=\{0,1,\ldots,p-1\}$, and let $f:X\to X$ be the shift operation $x\mapsto (x+1)\bmod p$. Clearly $f^p$ is the identity on$~X$. This function also gives an operation $\overline f:\mathcal P(X)\to\mathcal P(X)$ from the collection $\mathcal P(X)$ of $2^p$ subsets of $X$ to itself: $\overline f:S\mapsto\{\,f(x)\mid x\in S\,\}$.
The bit of group theory is that repeating $\overline f$ either gives a orbit of size $1$ or of size$~p$: for given $S$ either $\overline f(S)=S$, or else the $p$ subsets $S,\overline f(S),\overline f{}^2(S),\ldots,\overline f{}^{p-1}(S)$ are all different. Less formally: rotating a necklace of $p$ coloured beads by unit steps either gives just one colouring (for a monochrome necklace) or else $p$ different colourings. One also sees that $\overline f(S)=S$ only occurs when $S=\emptyset$ or $S=X$.
The subset $\overline f(S)$ of $X$ always has the same size as $S$. So iterating $\overline f$ partitions the $\binom pk$ subsets of size$~k$ into orbits of size$~p$, provided $k\notin\{0,p\}$ to exclude the fixed points $\emptyset$ and$~X$. This proves $p\mid\binom pk$.
You can also show the orbit property without group theory. Suppose some pair among the subsets $S,\overline f(S),\overline f{}^2(S),\ldots,\overline f{}^{p-1}(S)$ coincide: $\overline f{}^k(S)=\overline f{}^l(S)$ for $0\leq k<l<p$. Then because $\overline f$ is invertible one must have $\overline f{}^{l-k}(S)=S$; putting $d=l-k$ and iterating one gets that also $\overline f\,{}^{nd}(S)=S$ for $n=1,2,3,\ldots$. Since $d$ does not divide$~p$ and the exponents $nd$ can be reduced modulo$~p$ (as $\overline f\,{}^p$ is the identity), one of those powers reduces to $1$, so one concludes that $\overline f(S)=S$.
A: $\binom{p}{k} = \frac{p!}{k!(p-k)!} = \frac{p(p-1)...(p-k+1)}{k!}$
since $\binom{p}{k}$ is an integer, and none of the members of $k!$ can divide p ( since it's a prime), then $p|\binom{p}{k}$
A: There is a really nice way to phrase this, that should introduce you to some notation you should really know.
Let us define the function $v_p:\mathbb{Z}\to\mathbb{N}\cup\{\infty\}$ by defining $v_p(x)$ to be the highest $i$ such that $p^i$ divides $x$ (where we take $v_p(0)=\infty$). Let us then extend $v_p$ to a map $v_p:\mathbb{Q}\to\mathbb{Z}$ by setting $v_p\left(\frac{a}{b}\right)=v_p(a)-v_p(b)$. One can quickly check that $v_p$ enjoys the following nice property: 
$$v_p(xy)=v_p(x)+v_p(y)\quad\mathbf{(1)}$$
Moreover, we see by mere definition, that $p\mid x$ for $x\in\mathbb{Z}$ if and only if $v_p(x)>0$. Now, note that by $\mathbf{(1)}$ we have that  
$$\displaystyle \begin{aligned}v_p\left({p\choose k}\right) &= v_p\left(\frac{p!}{k!(p-k)!}\right)\\ &= v_p(p!)-v_p(k!)-v_p((p-k)!)\end{aligned}\quad\mathbf{(2)}$$
But, since $\ell!=1\cdots \ell$ we can use $\mathbf{(1)}$ again to deduce that for each $\ell\in\mathbb{N}$ one has that 
$$v_p(\ell!)=\sum_{j=1}^{\ell}v_p(j)$$
Now, if $j<p$ then evidently $p\nmid j$ so that $v_p(j)=0$. Thus,
$$v_p(k!)=\sum_{j=1}^{k}v_p(j)=\sum_{j=1}^{k}0=0$$
and
$$v_p((p-k)!)=\sum_{j=1}^{p-k}v_p(j)=\sum_{j=1}^{p-k}0=0$$
But
$$v_p(p!)=\sum_{j=1}^{p}v_p(j)=\sum_{j=1}^{p-1}v_p(j)+v_p(p)=\sum_{j=1}^{p-1}0+1=1$$
Thus, using $\mathbf{(2)}$ we may conclude that 
$$v_p\left({p\choose k}\right)=1-0-0=1$$
and thus $\displaystyle p\mid {p\choose k}$, and moreover $p$ is the highest power of $p$ dividing ${p\choose k}$.
A: By definition, $\binom{p}{k}$ can be expanded as follows: 
$\binom{p}{k}$ = $\frac{p!}{k!(p - k)!}$
Since $\binom{p}{k}$ is a positive integer number by definition, let's denote it as $n$, $n \in \mathbb N$. 
$n$ = $\frac{p!}{k!(p - k)!}$
Now, move the denominator to the left side: 
$n$ $\cdot$ $k!(p - k)!$ = $p!$
$p!$ is obviously divisible by $p$. 
Hence, $n$ $\cdot$ $k!(p - k)!$ is also divisible by $p$.
So, at least one factor must be divisible by $p$:

*

*$k!$ is not divisible by $p$ because $k < p$ and $p$ is prime.

*$(p - k)!$ is not divisible by $p$ because $(p - k) < p$ and $p$ is prime.

Thereof, $p \mid n \Rightarrow p \mid \binom{p}{k}$
A: Hint $\bmod p\!:\ (x\!+\!1)^p\!-x^p\!-1 \color{#c00}{=\, 0},\,$ having degree $ < p,\,$ and roots $\,1,2,\ldots, p\:$ by little Fermat.
Remark $ $ We used the well-known fact that a $\rm\color{#c00}{nonzero}$ polynomial over a field (or domain) has no more roots than its degree, so the above polynommial must be the $\rm\color{#c00}{zero}$ polynomial. More generally see Freshman's Dream Binomial Theorem / Frobenius endomorphism.
A: Let $N=\binom{p}{k}=\frac{p!}{k!(p-k)!}$. Then $N\cdot k!\cdot(p-k)!=p!$. Since $p$ clearly divides $p!$, it must also divide $N\cdot k!\cdot (p-k)!$. By Euclid's lemma, $p|N$ or $p|k!$ or $p|(p-k)!$. Since $k\in\{1,\dots,p-1\}$, it is impossible for $p$ to divide $k!$ or for $p$ to divide $(p-k)!$ (one can apply Euclid's lemma again to see this), and so $p$ must divide $N=\binom{p}{k}$.
A: Recall ${\displaystyle {p\choose k}=\frac{p!}{k!(p-k)!}=\frac{p\cdot p-1\cdot p-2\cdots 2\cdot1}{k\cdot k-1\cdots 2\cdot1\cdot p-k\cdot p-k-1\cdots2\cdot1}}$ and $a\mid b\Rightarrow b=a\cdot m$, where $m\in\mathbb{Z}$. Since $k\leq p-1$, it is obvious to assume $k<p$, i.e., $k!$ is contained somewhere in $p!$, so these terms will cancel out (honestly this step isn't even necessary, but it is nice to see expressions in lowest terms). The final term remaining in $p!$ will be $k+1$ in terms of $p$; denote this by $(k+1)^*$. This leaves us with $\frac{p\cdot p-1\cdots (k+1)^*}{p-k\cdot p-k-1\cdots2\cdot1}$. If we factor out $p$, then we obtain ${\displaystyle{p\choose k}=p\cdot\frac{p-1\cdots (k+1)^*}{p-k\cdot p-k-1\cdots2\cdot1}}$. Note that this holds since ${\displaystyle{p\choose k}}$ is an integer and $k!\nmid p$ (as mentioned above). Clearly $p$ divides ${\displaystyle{p\choose k}}$.
