$Gal(F(P)/F)$ is indeed a subgroup of $E[p^{N}]$. We see that it is not contained in $E[p^{N-1}]$, because otherwise $p^{N-1}P\in E(F)$. Thus, since $E[p^{n-1}]$ consists of all the elements of order less than $p^n$, order of $Gal(F(P)/F)$ is at least $p^N$. So we may take any $N \leq \log_p c$

$Gal(F(P)/F)$ is indeed a subgroup of $E[p^{n}]$. We see that it is not contained in $E[p^{n-1}]$, because otherwise $p^{n-1}P\in E(F)$. Thus, since $E[p^{n-1}]$ consists of all the elements of order less than $p^n$, order of $Gal(F(P)/F)$ is at least $p^N$. So we may take any $N \leq \log_p c$