Number of times $2^k$ appears in factorial 
For what $n$ does: $2^n | 19!18!...1!$?

I checked how many times $2^1$ appears:
It appears in, $2!, 3!, 4!... 19!$ meaning, $2^{18}$ 
I checked how many times $2^2 = 4$ appears:
It appears in, $4!, 5!, 6!, ..., 19!$ meaning, $4^{16} = 2^{32}$
I checked how many times $2^3 = 8$ appears:
It appears in, $8!, 9!, ..., 19!$ meaning, $8^{12} = 2^{36}$
I checked how many times $2^{4} = 16$ appears:
It appears in, $16!, 17!, 18!, 19!$ meaning, $16^{4} = 2^{16}$
In all, 
$$2^{18} \cdot 2^{32} \cdot 2^{36} \cdot 2^{16} = 2^{102}$$
But that is the wrong answer, its supposed to be $2^{150}$?

How many times $2$ divides the product $\prod_{i=1}^{19}i!$ ?
Let's call each term inside a factorial $i$. That way, $i = 1$ occurs in 19 factorials, $i = 2$ occurs in 18 factorials, and $i = 3$ occurs in 17 factorials etc.
$i = 2$ occurs 18 times. $1 \times 18 = 18$
$i = 4$ occurs 16 times. $2 \times 16 = 32$
$i = 6$ occurs 14 times. $1 \times 14 = 14$
$i = 8$ occurs 12 times. $3 \times 12 = 36$
$i = 10$ occurs 10 times. $1 \times 10 = 10$
$i = 12$ occurs 8 times. $2 \times 8 = 16$
$i = 14$ occurs 6 times. $1 \times 6 = 6$
$i = 16$ occurs 4 times. $4 \times 4 = 16$
$i = 18$ occurs 2 times. $1 \times 2 = 2$
$$18 + 32 + 14 + 36 + 10 + 16 + 6 + 16 + 2 = 150$$
