Big, bad, binary

Counting trailing zeroes is equal to counting powers of the base-- in this case, powers of 2.

In $n!$ , each even number $\leq n$ contributes a power of 2.

Each multiple of four contributes another power of 2.

Each multiple of eight contributes another power of 2.

And so on. Thus the number of factors of 2 in $n!$ is equal to $(\text{multiples of}\ 2^1,\ \leq n) + \cdots + (\text{multiples of}\ 2^i,\ \leq n) + \cdots \\ = \left\lfloor\frac n{2^1}\right\rfloor + \cdots + \left\lfloor\frac n{2^i}\right\rfloor + \cdots,$ where $2^i$ can be limited to be less than or equal to $n$ . The calculation becomes particularly easy when $n = 2^k$ , because the argument of the floor function $\lfloor\cdot\rfloor$ is a whole number in every instance:

$\text{factors of 2 in}\ (2^k)! = \frac{2^k}{2^1} + \cdots + \frac{2^k}{2^i} + \cdots + \frac{2^k}{2^k} \\ = 2^{k-1} + \cdots + 2^0 = 2^k-1.$

Substitute $k = 16$ and we find that there are $2^{16}-1 = \boxed{65535}$ factors of 2 in $(2^{16})!$ , and therefore 65335 trailing zeroes in its binary representation.

Let $n$ be a positive integer, let $p$ be a prime, and let $s$ be the sum of the digits in $n$ when expressed in base $p$ . Then the number of factors of $p$ in $n!$ is given by $\frac{n - s}{p - 1}.$

For $n = 2^{16}$ and $p = 2$ , $s = 1$ , so the number of factors of 2 in $(2^{16})!$ is simply $2^{16} - 1 = 65535$ .

As Arjen said, we just need to count the powers of 2 in $2^{16}$ . We can express this as

$\sum _{ n=1 }^{ 16 }{ \frac { { 2 }^{ 16 } }{ { 2 }^{ n } } } ={ 2 }^{ 16 }\sum _{ n=1 }^{ 16 }{ \frac { 1 }{ { 2 }^{ n } } }$

This is because every power of 2 less than or equal to 16 is a divisor of $2^{16}$ . On simplifying the GP, we get

${ 2 }^{ 16 }\left( 1-\frac { 1 }{ { 2 }^{ 16 } } \right) ={ 2 }^{ 16 }-1=65535$

In fact, with a little thought, it can also be proved that for any positive integral $n$ , ${ \left( { 2 }^{ n } \right) ! }_{ 2 }$ has $2^n - 1$ trailing zeroes in base 2.