TITLE: Why does $(128)!$ equal the product of these binomial coefficients $128! = \binom{128}{64}\binom{64}{32}^2 \dots \binom21^{64}$?
QUESTION [9 upvotes]: I'm working through some combinatorics practice sets and found the following problem that I can't make heads or tails of.
It asks to prove the following:
$$128! = \binom{128}{64}\binom{64}{32}^2\binom{32}{16}^4\binom{16}8^8\binom 84^{16}\binom 42^{32}\binom{2}{1}^{64}$$
Weird, huh? The first thing I noticed is that the exponents mirror the $r$ variables. I would normally just re-express each statement in $\frac{n!}{(n-r)!r!}$ form, but the exponents throw me for a loop. Are there any intuitions about factorials or nCr I should be considering here?

REPLY [10 votes]: Divide $128$ items in half, and assign one half a $1$ bit in the first digit and the other a $0$ bit. Then divide each half in half again, and in each half assign one half a $1$ bit in the second digit and the other a $0$ bit. Continue until the halves consist of single elements. Now each element has been assigned a binary number from $0$ to $127$. The left-hand side counts the number of ways of assigning the numbers, and the right-hand side counts the number of ways of performing the subdivisions.