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Show that if $p$ is a prime satisfying $n

The following is the attempted solution.

Let $n\in \mathbb{Z}$. Suppose $p$ is a prime such that $n

And since $p\in\{n+1,n+2,...,2n-1\}$ $\space$ we have $(n+1)(n+2)...(n+n)\equiv0 (mod\space p)$. That is $\frac{(2n)!}{n!}\equiv0 (mod\space p)$.

Therefore $\frac{n!}{n!}\frac{(2n)!}{n!}\equiv0 (mod\space p)$. But note that $g.c.d.(n!,p)=1$. So $\frac{(2n)!}{n!n!}\equiv0 (mod\space \frac{p}{g.c.d.(n!,p)})$. Hence the result.

Is the above solution correct? If not, can someone please give me a hint?

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    You have a correct idea that is central to a proof. Your finish is a little uneven, in that the $\gcd(n!,p) = 1$ could be developed earlier, and it would be useful in showing $\binom{2n}{n} \equiv 0 \bmod p$. – 2017-01-19
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    Yes I see it now. Thank you. :) – 2017-01-19

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I like the part about how all the entries in line $p,$ with $p $ PRIME, are divisible by $p$ except the endpoints. Divisiblity by $p$ is then automatic for all entries in a triangle descending and travelling inwards. Let me paste a picture and then identify an example...

Well, here is the easy one to see: look at all the entries in rows 5,6,7,8 that are divisible by $5,$ a kind of upside down triangle with apex at $70,$ which is $8$ choose $4.$

enter image description here

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    "Divisibility by $p$ is then automatic for all entries in a triangle descending and traveling inwards." I don't understand this part. Please explain. – 2017-01-19
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    @Janitha357 in row 5, entries 1,2,3,4 are 5, 10, 10, 5. In row 6, entries 2,3,4 are 15, 20, 15. In row 7, entries 3,4 are 35, 35. Finally, in row 8, entry 4 is 70. – 2017-01-19
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    @Janitha357 a similar argument shows that there are three consecutive entries in row 14, those being 1001, 2002, 3003. That is, all three are divisible by 7, 11, and 13. The 7 works because 14 = 2 * 7 and we are not using the central position, while 7 is an odd prime. – 2017-01-19
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    Thank you. This is really nice. – 2017-01-19