Suppose that $\,p,f,g\,$ are polynomials over $\,\Bbb C\,$ (or any field) and $\,a,\,a_i\in \Bbb C\,$ are distinct points.
Suppose $\ p(x) = f(x)/g(x)\ $ for $\ x = a_i,\ i = 1,\ldots, k = \deg(gp) > \deg f,\, $ and $\, g(a) = 0.$
Then $\ q\, := g\,p-f \:$ has degree $\,k\,$ and $\,k\,$ distinct roots $\,a_i$ so $\ q = c\,(x\!-\!a_1)\cdots (x\!-\!a_{k})\,=:\,c\,h $
for some constant $\,c.\ $ But $\,\ g(a)=0\,\Rightarrow\, -f(a) = q(a) = c\,h(a),\ $ so $\ c = -f(a)/h(a).$
Therefore $\ p\, =\, \dfrac{f+q}g\, = \,\dfrac{f}g - \dfrac{f(a)h}{g\, h(a)}\, =\, \dfrac{f}g\, -\, \dfrac{f(a)}g \dfrac{(x-a_1)\cdots(x-a_{k})}{ (a-a_1)\cdots(a-a_{k})}$
In OP: $\,\ \begin{eqnarray}f = x,\ \ g = x\!+\!1\ \ \\ a_i = i-1,\ a=-1\end{eqnarray}\ $ so $\,\ p = \dfrac{x}{x+1} +\, \dfrac{1}{x+1}\, \dfrac{\,\ x\ \,\cdots\,\ (x\,-\,n)}{ -1\,\cdots(-1-n)}=\smash{\dfrac{x+(-1)^{n+1}{x\choose n+1}}{x+1}}$
Therefore, evaluating the prior at $\ x = n+1\ $ we deduce that $\ p(n+1)\, =\, \smash{\dfrac{n+1+(-1)^{n+1}}{n+2}}$