A power of 2 meets a factorial

If we consider $500!,$ we can calculate how many of the terms in the product are divisible by $2^k$ for each value of $k.$

$\begin{array}{|r|cccccccc|} \hline k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\ \hline \mbox{num} & 250 & 125 & 62 & 31 & 15 & 7 & 3 & 1\\ \hline \end{array}$

This means that $2^{250 + 125 + 62 + 31 + 15 + 7 + 3 + 1} = 2^{494}$ is the largest power of dividing $500!$ We can factor the next few even numbers $502 = 2 \times 251, 504 = 8 \times 63, 506 = 2 \times 253, 508 = 4 \times 127.$ This tells us that $2^{499}$ divides $506!$ (also $507!$ ) and $2^{501}$ divides $508!$