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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

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