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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 $p$

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up vote 2 down vote accepted

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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