<s>[INST] 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 $2^{150}$?
 [/INST] It might be more helpful to do this recursively.
Let $T(n) = \prod_{k=1}^n k!$.
We will use the notation: $2^{r} \| m$ to mean that $2^r$ is the largest power of $2$ that divides $m$.
Then we have $2 \| 2! = T(2)$. We also know that $2 \| 3!$, so $2^2 \| T(3) = 3! T(2)$. Continuing:
$$2^3 \| 4!$$
$$2^3 \| 5!$$
$$2^4 \| 6!$$
$$2^4 \| 7!$$
$$2^7 \| 8!$$
$$2^7 \| 9!$$
$$2^8 \| 10!$$
$$2^8 \| 11!$$
$$2^{10} \| 12!$$
$$2^{10} \| 13!$$
$$2^{11} \| 14!$$
$$2^{11} \| 15!$$
$$2^{15} \| 16!$$
$$2^{15} \| 17!$$
$$2^{16} \| 18!$$
$$2^{16} \| 19!$$
If we take the sum of all of those powers, $$2\cdot 1 + 2\cdot 3 + 2\cdot 4 + 2\cdot 7 + 2\cdot8 + 2 \cdot 10 + 2 \cdot 11 + 2 \cdot 15 + 2 \cdot 16$$
$$=2(1+3+4+7+8+10+11+15+16) = 2(75) = 150.$$
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