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

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.

Prove n<2^{n}


(1): P_{1} is true, since 1<2^{1}

(2): Assume P_{k} is true: k<2^{k}.

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

From P_{k}, we see that 2(k)<2(2^{k})=2^{k+1} and conclude that k+1<2^{k+1}.

Thus, hence, therefore, P_{k+1} is true and QED.