A polynomial that equals k/(k+1) at 0, 1, …, n

Auxiliary polynomial. $Q(x)=(x+1)P(x)-x$ has degree at most $n+1$, and $Q(k)=(k+1)\frac k{k+1}-k=0$ for $k=0,\dots,n$. These are $n+1$ distinct roots, so
$$Q(x)=c\,x(x-1)\cdots(x-n)$$
for some constant $c$.

The constant. At $x=-1$: $Q(-1)=0\cdot P(-1)+1=1$, while the right side is
$$c(-1)(-2)\cdots(-1-n)=c\,(-1)^{n+1}(n+1)!.$$
Therefore $c=\dfrac{(-1)^{n+1}}{(n+1)!}$.

At $x=n+1$.
$$(n+2)P(n+1)-(n+1)=c\,(n+1)!=(-1)^{n+1},$$
so
$$P(n+1)=\frac{n+1+(-1)^{n+1}}{n+2}.$$
In particular $P(n+1)=1$ for $n$ odd and $\dfrac{n}{n+2}$ for $n$ even.

Check. $n=2$: the polynomial through $(0,0),(1,\frac12),(2,\frac23)$ takes the value $\frac12$ at $3$, and the formula gives $\frac{3-1}4=\frac12$ ✓.

Lesson. When values of $P$ are given on $n+1$ points, build an auxiliary polynomial vanishing there; its degree forces its shape up to a constant.