求大神解一下这道题，谢谢，要个过程的
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For n=0, left=2^0=1; right=1+1=2. 2>=2. So the argument is right.
Assume that the argument is right when n=k, where k is a natural number. 2^k>=k+1.
When n=k+1. 2^(k+1)=2*2^k>=2*(k+1)=2k+2=k+(k+2)>=k+2. So when n=k+1, 2^(k+1)>=(k+1)+1 is approved. Based on mathematical induction, for all natural numbers n, we have 2^n>=n+1.
Assume that the argument is right when n=k, where k is a natural number. 2^k>=k+1.
When n=k+1. 2^(k+1)=2*2^k>=2*(k+1)=2k+2=k+(k+2)>=k+2. So when n=k+1, 2^(k+1)>=(k+1)+1 is approved. Based on mathematical induction, for all natural numbers n, we have 2^n>=n+1.
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