I wish to prove the following: Let $g$ be a polynomial with integer coefficients and let $$h(x)=\frac{g(x)}{(p-1)!}$$ for some $p\in\Bbb{N}$.
Then $\forall i\geq p:h^{(i)}$ is a polynomial with integer coefficients such that each is divisible by $p$. My go: For $p>\deg{g}$ proof is trivial because $h^{(p)}=0$ and for any $i\geq p$ too. Let $p\leq \deg{g}$. Formally, I want to prove by induction, prove the base step and then induction holds, because taking more derivatives doesn't change the divisibility.
So, i started writing out the derivatives... $$h(x)=\frac{g_0}{(p-1)!}+\frac{g_1x}{(p-1)!}+\frac{g_2x^2}{(p-1)!}+\frac{g_3x^3}{(p-1)!}+\frac{g_4x^4}{(p-1)!}+\cdots+\frac{g_nx^n}{(p-1)!}$$ $$h'(x)=\frac{g_1}{(p-1)!}+\frac{2g_2x}{(p-1)!}+\frac{3g_3x^2}{(p-1)!}+\frac{4g_4x^3}{(p-1)!}+\frac{5g_5x^4}{(p-1)!}+\cdots+\frac{ng_nx^{n-1}}{(p-1)!}$$ $$h''(x)=\frac{2\cdot 1g_2}{(p-1)!}+\frac{3\cdot 2g_3x}{(p-1)!}+\frac{4\cdot3g_4x^2}{(p-1)!}+\frac{5\cdot4g_5x^3}{(p-1)!}+\frac{6\cdot 5g_6x^4}{(p-1)!}+\cdots+\frac{n(n-1)g_nx^{n-2}}{(p-1)!}$$
For some $k$-th derivative the term with $g_{k-1}$ vanishes and the absolute term has $k!$ in the numerator: $$h^{(p)}(x)=\frac{p!g_p}{(p-1)!}+\frac{\frac{(p+1)!}{1!}g_{p+1}x}{(p-1)!}+\frac{\frac{(p+2)!}{2!}g_{p+2}x^2}{(p-1)!}+\frac{\frac{(p+3)!}{3!}g_{p+3}x^3}{(p-1)!}+\cdots$$ Now in the first term $p!$ and $(p-1)!$ cancels out and we are left with $pg_p$ which is divisible by $p$.
Is this reasoning for the next terms correct?
For some $j$-th term in $h^{(p)}(x)$ we cancel out the $(p-1)!$ and in the numerator we are left with $\frac{(p+j)(p+j-1)(p+j-2)\cdots (p+2)(p+1)p}{j!}g_{p+j}x^{j}$. Now $p+j,p+j-1,p+j-2...p+2,p+1$ is sequence of $j$ consecutive integers, thus for any $1\leq i\leq j$ (thus all factors of $j!$) we find some number $l$ in the sequence such that $i \mid l$.
I guess you want to show that the product of $p$ consecutive positive integers is divisible by $p!.$ Consider that $$ \frac{N(N-1)\dots(N-p+1)}{p!}={N\choose p}$$