Algebra and polynomials Let $p(x)$ be a polynomial with degree $2008$ and leading coefficient $1$, such that, $p(0) = 2007; p(1) = 2006; p(2) = 2005 ... p(2007) = 0$. 
Determine $p(2008)$.
I found that the constant is $2007$ but I am not able to find anything substantial other than that. I think that it is $-1$ just by virtue of the pattern forming, however, I am not able to prove this. Also, the answers are not given so I have no way of knowing if I am right. 

If you try LaGrange interpolation your answer will simplify to $p(x) = 2007 - x$.  This is the smallest degree polynomial going through each point.  However, the question requires that the polynomial has degree 2008.
To fix this, consider $q(x) = p(x) - (2007-x)$.  The requirements on $p(x)$ imply $q(x)$ has zeros at $x=0, 1, ..., 2007$.  The only monic polynomial of degree 2008 with these zeros is$q(x) = x(x-1)(x-2)...(x-2007).$ Hence
$$p(x) = x(x-1)(x-2)...(x-2007) + (2007-x).$$
Thus, $p(2008) = 2008! - 1$.
