1. ## mathematical induction again

I need to use mathematical induction to prove $\displaystyle 2^n<n!$ for all n>=4. The base case is easy, and in the inductive step I get to $\displaystyle 2(2^n)<(n+1)n!$ But am I going in the wrong direction, or am I just missing something?

2. Assume $\displaystyle 2^k < k!$ for $\displaystyle k \geq 4$. Then $\displaystyle 2^{k+1} < 2k! < (k+1)! = k!(k+1)$. $\displaystyle k+1 > 2$ for $\displaystyle k >1$. Hence $\displaystyle 2^{k+1} < (k+1)!$ for $\displaystyle k \geq 4$ and we have proved the inductive step.

3. Sorry, stupid typo
$\displaystyle 2^n<n!$ for all
$\displaystyle n>=4$

4. Assume that $\displaystyle N \ge 4\quad \& \quad 2^N < N!$ for the inductive step.

Then $\displaystyle 2^{N + 1} = 2\left( {2^N } \right) < \left( {N + 1} \right)\left( {N!} \right) = \left( {N + 1} \right)!$

5. The answer to these things is always staring me right in the face, and I just can't see it until someone points it out. Thanks again

6. Originally Posted by tottenc
The answer to these things is always staring me right in the face, and I just can't see it until someone points it out. Thanks again
It just takes some practice.

-Dan