Find the value of the polynomial at a given point f is a polynomial of degree $1007$ ,if $f(k) = 2^k$ for $0\le k \le 1007$ find the value of $f(2015)$
The solution should be $f(2015) = 2^{2014}$ and the polynomial $p(x) = \sum_{k=0}^{1007}{x \choose k}$ but i don't get how we can arrive to that.
 A: The idea is to use the identity
$$
{{n}\choose{0}}+ {n \choose 1} + \cdots + {n \choose n} = 2^n
$$
to create a polynomial of degree 1007 that has the desired properties.  We can see that this is true by the binomial theorem. Observe that
$$
(x+y)^n = \sum_{k=0}^{n}{n \choose k} x^{n-k} y^{k},
$$
so letting $x = y =1$ we obtain the desired result.
With this result in mind, we can show the polynomial 
$$
p(x) = \sum_{k=0}^{1007} {x \choose k} 
$$
satisfies the desired properties. Furthermore, since we are specifying 1008 points that a 1007 degree polynomial maps to, this must be the unique polynomial of this degree that does this. Finally, calculating $p(2015)$ requires a one last binomial coefficient identity. For any $n$ and $k$,
$$
{n \choose k} = {n \choose n-k}.
$$
Using this and the previous identity, we see that for any Odd $n = 2n' + 1$,
\begin{align*}
2^n &= \sum_{k=0}^{n} {n \choose k} \\
& = \sum_{k=0}^{n'} {n \choose k} + \sum_{k=n'+1}^{n} {n \choose k} \\
& = \sum_{k=0}^{n'} {n \choose k} + \sum_{k=n'+1}^{n} {n \choose n - k}.
\end{align*}
The latter sum is equal to ${n \choose n'} + {n \choose n' -1 }+ ... + {n \choose 0} $, while the former is equal to ${n \choose 0} + ... + {n \choose n'}$. Therefore, they are equal, so
\begin{align*}
2^n &= 2 \sum_{k=0}^{n'} {n \choose k} \\
& \text{and} \\
2^{n-1} &=\sum_{k=0}^{n'} {n \choose k}.
\end{align*}
This identity allows us to calculate $p(2015)$. Observe that $2015 = 2 \times 1007 +1$. Then by the above identity,
\begin{align*}
p(2015) & = \sum_{k=0}^{1007} {2015 \choose k} \\
& =  2^{2015-1} \\
&= 2^{2014},
\end{align*}
as desired.
A: Hint $\ $ Write $\,f(x) = c_0 + c_1 x + c_2 x(x-1) + c_3 x(x-1)(x-2) + \cdots$ then successively evaluate at $\,x=0,1,2\ldots$ to deduce $\,c_n = 1/n!\,\ $ (this is essentially Newton interpolation).
