TITLE: exercise of derivates and divisibility
QUESTION [4 upvotes]: How to prove that if $p$ is a prime number for all $i\geq p$ and $k\geq 0$ the coefficient of $$\frac{d^i}{dx^i}\left(\frac{x^{p+k}}{(p-1)!}\right)$$ is a integer number multiple of $p$

REPLY [2 votes]: We know, $$\frac{d^n (x^m)}{dx^n}=m(m-1)(m-n+1)x^{m-n}$$
So,  $$\frac{d^i}{dx^i}\left(\frac{x^{p+k}}{(p-1)!}\right)=\frac {(p+k)(p+k-1)\cdots(p+k-i-1)}{(p-1)!}x^{p+k-i}$$
So, the coefficient of $x^{p+k-i}$ is $$\frac {(p+k)(p+k-1)\cdots(p+k-i-1)}{(p-1)!}$$ 
As $i\ge p,$ if $i=p,$ the coefficient becomes $$\frac {(p+k)(p+k-1)\cdots(k-1)}{(p-1)!}=\binom {p+k}{k-1}$$ which is an integer.
The denominator$(p-1)!$ is co-prime with $p$.
The numerator contains $p+k-(k-1)+1=p+2$ consecutive terms hence is divisible by $p$
So, the coefficient is divisible by $p$.
On further differentiation i.e., for the higher values of $i$ , some integers (till the multiplier becomes $0$)will be multiplied with the coefficient, hence the later would remain divisible by $p$.
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