For $P(x)$ a degree-$n$ polynomial such that $P(k)=\frac{k}{k+1}$ for $k=0,1,2,\ldots,n$, find $P(n+1)$. The problem, from USAMO $1975$, reads:

If $P(x)$ denotes a polynomial of degree $n$ such that $P(k) = \frac{k}{k+1}$ for $k = 0,1,2, \ldots, n$. Find $P(n+1)$.

My attempt:
We can represent $p(x)$ as another polynomial which would satisfy all the given conditions. This polynomial would be:
$P(x)$ = $x(x-1)(x-2).....(x-n) + \frac{x}{x+1}$
In other words,
$\frac{P(x)(x+1)-x}{(x+1)}$ = $x(x-1)(x-2)....(x-n)$, or:
${P(x)(x + 1)-(x)}$ = $x(x^2$ $- 1)(x - 2)......(x-n)$
now, simply putting $x = n + 1$, we have
${P(n+1)(n+2)-(n+1)} = \frac{(n + 1)(n^2+ 2n)(n - 1)!}{2}$           (after some simplification)
thus, our final expression would look something like:
$P(n + 1) = \frac{(n + 1)(n^2+ 2n)(n - 1)!+2(n+1)}{2(n + 2)}$
which is not the final answer.

So $$Q(x) = (x+1)P(x)-x$$ is of degree $n+1$ and has zeroes $0,1,2,...,n$, thus we can write it $$Q(x) = ax(x-1)(x-2)\cdots (x-n)$$ so $$(x+1)P(x) =ax(x-1)(x-2)\cdots (x-n)+x$$ which is valid for all $x$ and in particulary for $x=-1$: $$0 =a(-1)^{n+1}(n+1)!-1$$ and for $x=n+1$ we get: $$(n+2)P(n+1) =a(n+1)!+n+1$$ so $$P(n+1)= {(-1)^{n+1} +n+1\over n+2}$$
