(a) (2^n) ≤ n! , n≥4

Base Step: sub in n=1 and yes, it works!

Inductinve step: assume (2^n) ≤ n! and show (2^(k+1)) ≤ (k+1)! ,K≥4 holds.

(2^(k+1)) ≤ (k+1)!

(2)

**(2^k) ≤ (K!)**(K+1)

So the bold part is the original equation. In the above, the left side was multiplied by 2 and the right side was multiplied by (K+1). So, anything multiplied by 2 is less than anything multiplied by (K+1), K≥4. Is this proof reasonable?

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I initially tried this:

(2^(k+1))

= 2(2^K)

= ???

= ???

≤ (k+1)!

What I am trying to show is to work my way from (2^(k+1)) by expanding it and eventually show that it is less that (K+1)!

somebody care to comment?

Thanks