6. [5] Let $p_{0}(x), p_{1}(x), p_{2}(x), \ldots$ be polynomials such that $p_{0}(x)=x$ and for all positive integers $n$, $\frac{d}{d x} p_{n}(x)=p_{n-1}(x)$. Define the function $p(x):[0, \infty) \rightarrow \mathbb{R} x$ by $p(x)=p_{n}(x)$ for all $x \in[n, n+1]$. Given that $p(x)$ is continuous on $[0, \infty)$, compute
$$
\sum_{n=0}^{\infty} p_{n}(2009)
$$
Answer: $e^{2010}-e^{2009}-1$
Solution: By writing out the first few polynomials, one can guess and then show by induction that $p_{n}(x)=\frac{1}{(n+1)!}(x+1)^{n+1}-\frac{1}{n!} x^{n}$. Thus the sum evaluates to $e^{2010}-e^{2009}-1$ by the series expansion of $e^{x}$.