I'm having issues proving that for every n = 4, n! = 2^n using induction

Question

I'm having issues proving that for every n >= 4, n! >= 2^n using induction

zepdrix · Answer

$$n=4 \rightarrow 4! \ge 2^4$$$$1*2*3*4 \ge 2*2*2*2$$$$24 \ge 16$$
If n=k
$$k! \ge 2^k$$$$1*2*3*4*....*(k-1)*(k) \ge 2_1 *2_2 *2_3* 2_4* ... *2_{k-1} *2_k$$


Blahhh I don't remember induction. It seems like, from here, you can conclude that since the product of the first 4 terms is greater than 2^4, and EVERY other term is greater than 2, the remaining product will be greater than 2^k.

I dunno sorry I forget how to set these up :P Hopefully someone can help ya..

zepdrix · Answer

@TuringTest

turingtest · Answer

$$k!\ge2^k\implies\frac{2^k}{k!}\le1$$now for $n=k+1$ we have$$(k+1)!\ge2^{k+1}$$$$k!(k+1)\ge2(2^k)$$$$k+1\ge2\left(\frac{2^k}{k!}\right)$$now because we know that $$\frac{2^k}{k!}\le1$$we know that$$2\left(\frac{2^k}{k!}\right)\le2$$and we know that k+1 is at least 5 since k is at least 4, so$$ k+1>2\ge2\left(\frac{2^k}{k!}\right)$$I think that is a solid proof