$\tbinom{2p}{p}-2$ is divisible by $p^3$ The problem is as follows:
Let $p>3$ be a prime. Show that $\tbinom{2p}{p}-2$ is divisible by $p^3$. The only thing I can think of is that $(2p)!-2(p!)^2$ is divisible by $p^2$ which doesn't help me much. Can someone point me in the right direction? Is there a combinatorial approach to this problem?
Thanks
 A: $${2p\choose p}=\frac{(2p)(2p-1)\ldots (2p-(p-1))}{p!}=\frac{2(2p-1)\ldots (2p-(p-1))}{(p-1)!}=2{2p-1\choose p-1}$$
Now by Wolstenholme's theorem
$${2p\choose p}\equiv 2\cdot1\equiv 2\mod p^3$$
${} {} {}$
A: Wolstenholme's Theorem tells us that
$$\binom {2p-1}{p-1}=1\pmod{p^3}$$
and from here...
A: $\tbinom{2p}{p} = \frac{(p+1)(p+2)...(p+p-1)(p+p)}{1.2...(p-1)p} = \frac{(p+1)(p+2)...(p+p-1)2}{1.2...(p-1)}$ 
$\tbinom{2p}{p} -2 $ will be divisible by $p^3$ 
iff $ \frac{(p+1)(p+2)...(p+p-1)2}{1.2...(p-1)} -2 $ is divisible by $p^3$
iff $ ((p+1)(p+2)...(p+p-1)) -(p-1)! $ is divisible by $p^3$  as (2,p)=1 and (p,(p-1)!)=1
Let f(x)= $\prod_{1≤r≤p-1}(x+r) = \sum _{0≤r≤p-1}a_rx^r$ .
Then (x+1)f(x+1)=(x+p)f(x).
Putting the values  of f(x) and  f(x+1) and comparing the coefficients of the different powers of x,
for $x^p$, 1=1
for $x^{p-1}, p+a_{p-1}=pC_1+a_{p-1}$ 
and so on.
Clearly,p |$a_r$  and by Wolstenholme's Theorem  $p^2|a_1$  
So in $\prod_{1≤r≤p-1}(p+r)$,
$a_0$ is (p-1)!,
the co-efficient of p($a_1$) is the sum of product of r taken 2 at a time,
the co-efficient of $p^2$($a_2$) is the sum of product of r taken 3 at a time and the rest terms are divisible by $p^3$.
As $p|a_2$ and $p^2|a_1$, the result follows.
