n^n≥(n+1)^(n-1),當(dāng)n=1時(shí),左邊=1,右邊=1,上式成立；當(dāng)n=2時(shí),左邊=4,右邊=3,上式成立；設(shè)當(dāng)n=k時(shí),上式成立,k^k≥(k+1)^(k-1),兩邊取對數(shù)得：klogk≥(k-1)log(k+1),log(k+1)+klogk≥klog(k+1),klog(k+1)+log(k+1)≥2klog(k+1)-klogk,左邊=(k+1)log(k+1)=log[(k+1)^(k+1)],右邊=k[log(k+1)²-logk]=klog[(k+1)²/k]=klog[(k+2)+1/k]>[(k+1)-1]log[(k+1)+1]=log[(k+1)+1]^[(k+1)-1],則log[(k+1)^(k+1)]>log[(k+1)+1]^[(k+1)-1],[(k+1)^(k+1)]>[(k+1)+1]^[(k+1)-1],即當(dāng)n=k+1時(shí),上式成立,n^n≥(n+1)^(n-1).