We know that the $(n+1)^{th}$ forward finite difference of $P$ evaluated at $x=0$ is \[0 = \Delta^{n+1}[P](0) = \sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^{n+1-k}P(k)=P(n+1)+\sum_{k=0}^{n}(-1)^{n+1-k},\]so \[P(n+1)=\sum_{k=0}^{n}(-1)^{n-k}=2\left\{\frac{n+1}{2}\right\}.\]

For $k=0, 1, \ldots , n$, let $w_k = k$ and \[c_k=\dbinom{n+1}{k}^{-1}=\frac{k!(n+1-k)!}{(n+1)!}.\] Define $f_0, f_1, \ldots f_n$ as \[f_i(x)=\prod_{j=0}^n (x-w_j).\] Hence we have \[f_k(k)=(-1)^{n-k}k!(n-k)!\] and \[f_k(n+1)=\frac{(n+1)!}{(n+1-k)}.\] By Lagrange interpolation formula we have \[P(n+1)=\sum_{k=0}^nc_k\frac{f_k(n+1)}{f_k(k)}=\sum_{k=0}^n(-1)^{n-k},\] which is $0$ if $N$ is odd and $1$ if $n$ is even.