Find $f(2015)$ given the values it attains at $k=0,1,2,\cdots,1007$ 
Let $f$ be a polynomial of degree $1007$ such that $f(k)=2^k$ for $k=0,1,2,\cdots, 1007$. Determine $f(2015)$.

Taking $f(x)=\sum_{n=0}^{1007} a_n x^n $, (whence $a_0=1$), I tried to combine $$a_1=1-\sum_{n=2}^{1007} a_n$$ and the easily derivable $$ a_n = \frac{2^n-1}{n^n} - \sum_{k=1}^{n-1} a_k n^{k-n} - \sum_{k=n+1}^{1007} a_k n^{k-n},$$ hoping to find several null coefficients, but that didn't work. I looked for the solution on Yahoo Answers, and I came to know it's about binomial coefficients, Tartaglia in particular; however the explanation is confusing to me, I'd like a better one, or even a different approach.
 A: It is not difficult to guess what is the representation of $f(x)$ with respect to the binomial base. 
For fearless people that do not like to guess, Lagrange's interpolation gives:
$$ f(x) = \sum_{j=0}^{1007} 2^j\cdot\,\!\!\!\!\! \prod_{\substack{k\in[0,1007]\\k\neq j}}\frac{(x-k)}{(j-k)}=\sum_{j=0}^{1007}\frac{2^j}{x-j}\cdot 1008!\binom{x}{1008}\cdot\frac{(-1)^j}{j!(1007-j)!} \tag{1}$$
hence:
$$ f(2015)=\sum_{j=0}^{1007}\frac{(-2)^j\cdot 1008}{2015-j}\binom{2015}{1008}\binom{1007}{j}$$
or:
$$ f(2015)=2^{1007}\binom{2015}{1008}\sum_{j=0}^{1007}\frac{(-1)^j\cdot 1008}{2^j\cdot(1008+j)}\binom{1007}{j}, \tag{2}$$
but:
$$\begin{eqnarray*}\sum_{j=0}^{1007}\frac{(-1)^j}{2^j(1008+j)}\binom{1007}{j}&=&2^{1008}\int_{0}^{1/2}x^{1007}(1-x)^{1007}\,dx\\&=&2^{1007}\frac{\Gamma(1008)\Gamma(1008)}{\Gamma(2016)}\\&=&\frac{2^{1007}}{1008}\cdot\frac{1}{\binom{2015}{1008}}\tag{3}\end{eqnarray*} $$
by the Euler beta function, so line $(3)$ nicely simplifies to:

$$ f(2015)=2^{2014}.\tag{4} $$

A: Hint: $f(x)=\sum_{r=0}^{1007}\,\binom{x}{r}$, where $\binom{x}{r}:=\frac{x(x-1)\cdots(x-r+1)}{r!}$ for every positive integer $r$ and $\binom{x}{0}:=1$, satisfies the required condition.  Show that it is the only one satisfying this property.  Then, verify that $f(2015)=2^{2014}$.
A: You are basically supposed to guess that since $f(x) \approx 2^x = (1 + 1)^k = \sum_{k=0}^{x} {x \choose k} 1^k 1^{x - k}$, that perhaps you can crop the series at $1008$ terms to get $f(x) = \sum_{k = 0}^{1007} {x \choose k}$.
Then use the fact that ${x \choose k} = {x \choose x - k}$ to get $$\sum_{k = 0}^{1007} {x \choose k} = \frac 12 \sum_{k = 0}^{2015} {x \choose k}$$
and apply the binomial sum again with 
$$\sum_{k = 0}^{2015} {x \choose k} = \sum_{k = 0}^{2015} {x \choose k} 1^k 1^{2015 - k} = (1 +1)^{2015}$$
