We'll use induction on $d$.base cases are trivial.Note that $Q(x) = P(x+1)-P(x)$ is good and has degree $d-1$ and leading coefficient $da_d$ where $a_d$ is the leading coefficient of $P$. So by induction, $N_{d-1}da_d$ is integer and we're done in the where $d$ is composite. So assume $d$ is a prime number. Since $P(1)$ is integer , sum of the coefficient of $P$ is an integer and ftsoc $a_d$ has factor $d$ in it's denominator(and it's a rational number indeed by LIF on arbitrary $d$ integer points of $P$) . So there's some $i<d$ st $a_i$ has $d$ in it's denominator. But then , the coefficient $x^{i-1}$ in $P'$ would have $d$ in it's denominator. But it has degree $d-1$ and it's integer at integer points which is well-known that it can be shown in the form : $$\sum b_i \binom {X}{i}$$Where $b_i$ is an integer for each $i \leq d-1$. So $(d-1)!P'$ has integer coefficients but the coefficient of $x^{i-1}$ isn't an integer which is impossible and we're done.

Solved with starchan, mueller.25, AdhityaMV We induct on the value of $d$. If $d$ is composite, then consider the polynomial $P(x+1) - P(x)$, it has degree $d-1$ and has derivative to also be integer at integer points. So by induction, $N_{d-1} \cdot da_d = N_d a_d$ is an integer, as desired. Now assume $d = p$ is prime. Then if the leading coefficient has negative $\nu_p$, then clearly some other coefficient must also have negative $\nu_p$, otherwise it cannot always be an integer. But then this means $P'(x)$ also has a coefficient with negative $\nu_p$, but this is impossible as if $P'(x)$ only takes integer values, then $(p-1)! \cdot P'(x)$ is an integer coefficient polynomial. So the leading coefficient does not have denominator divisible by $p$, so the induction works out anyway, so we're done. $\blacksquare$

alinazarboland wrote: We'll use induction on $d$.base cases are trivial.Note that $Q(x) = P(x+1)-P(x)$ is good and has degree $d-1$ and leading coefficient $da_d$ where $a_d$ is the leading coefficient of $P$. So by induction, $N_{d-1}da_d$ is integer and we're done in the where $d$ is composite. So assume $d$ is a prime number. Since $P(1)$ is integer , sum of the coefficient of $P$ is an integer and ftsoc $a_d$ has factor $d$ in it's denominator(and it's a rational number indeed by LIF on arbitrary $d$ integer points of $P$) . So there's some $i<d$ st $a_i$ has $d$ in it's denominator. But then , the coefficient $x^{i-1}$ in $P'$ would have $d$ in it's denominator. But it has degree $d-1$ and it's integer at integer points which is well-known that it can be shown in the form : $$\sum b_i \binom {X}{i}$$Where $b_i$ is an integer for each $i \leq d-1$. So $(d-1)!P'$ has integer coefficients but the coefficient of $x^{i-1}$ isn't an integer which is impossible and we're done. Can you explain in detail why we have such form? $$\sum b_i \binom {X}{i}$$