Find all positive integers such that 2^n < n!

Claim: $2^n < n!$ holds for all positive integers $n \geq 4$ , and no others.
Proof. To show that it does not hold for $n=1,2,3$ we compute:

$2^1=2 \not< 1!=1, \qquad 2^2 = 4 \not< 2! = 2, \qquad 2^3 = 8 \not< 3! = 6.$

Now, for the case $n=4$ on the left we have

$2^n = 2^4 = 16.$

While, on the right we have,

$n! = 4! = 24.$

Since 16 is indeed less than 24, we the inequality holds for $n =4$ . Assume then that the inequality holds for some $n =k \in \mathbb{Z}_{\geq 4}$ . Then,

$2^k < k! \ \implies \ (2^k)(k+1) < (k!)(k+1) \ \Righarrow \ 2^{k+1} < (k+1)!.$

Where the last inequality uses the fact that since $k \geq 4 > 2$ so $2^k (k+1) > (2^k)\cdot2 = 2^{k+1}$ . Thus, the inequality holds for the case $k+1$ ; and hence, for all $n \in \mathbb{Z}_{\geq 4}. \qquad \blacksquare$

Stumbling Robot

Find all positive integers such that 2^n < n!

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