[dao]

how do I prove by mathematics induction with this problem
2^n>n

[worthbeads]

Try using examples. For instance, if N=5, 2^5=32 and 32>n. It would work for a negative number too. If N=-2, 2^-2=1/4 and 1/4>-2. It works for zero. If N=0 then 2^0=1 and 1>0. SO if it works for positive numbers, negative numbers, and zero it should work, right? That's the way I would do it. There is probably a better way to do it.

[galactus]

Prove


(1): is true, since

(2): Assume is true: .

Now, k+1<k+k=2(k) for k>1.

From , we see that and conclude that .

Thus, hence, therefore, is true and QED.