Let positive integers $m\le k \le n$. Show that $$\binom{n}{k}\binom{k}{m} =\binom{n}{m}\binom{n-m}{k-m} =\binom{n}{k-m}\binom{n-k+m}{m}$$

Lizzie writes a list of fractions as follows. First, she writes $\frac{1}{1}$ , the only fraction whose numerator and denominator add to $2$. Then she writes the two fractions whose numerator and denominator add to $3$, in increasing order of denominator. Then she writes the three fractions whose numerator and denominator sum to 4 in increasing order of denominator. She continues in this way until she has written all the fractions whose numerator and denominator sum to at most $1000$. So Lizzie’s list looks like: $$\frac{1}{1}, \frac{2}{1} , \frac{1}{2} , \frac{3}{1} , \frac{2}{2}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2} , \frac{2}{3}, \frac{1}{4} ,\cdots, \frac{1}{999}$$

Let $p_k$ be the product of the first $k$ fractions in Lizzie’s list. Find, with proof, the value of $p_1 + p_2 +\cdots+ p_{499500}$.

Show that $$\binom{n}{1}-\frac{1}{2}\binom{n}{2}+\frac{1}{3}\binom{n}{3}-\cdots+(-1)^{n+1}\binom{n}{n}= 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$$