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.

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