由 b≤2p−2b\le2p-2b≤2p−2，得 ap=b!+p≤(2p−2)!+p<p2pa^p=b!+p\le(2p-2)!+p<p^{2p}ap=b!+p≤(2p−2)!+p<p2p（利用粗略估计 (2p−2)!=∏k=1p−1k(2p−1−k)<(p(p−1))p−1<p2p(2p-2)!=\prod_{k=1}^{p-1}k(2p-1-k)<(p(p-1))^{p-1}<p^{2p}(2p−2)!=∏k=1p−1k(2p−1−k)<(p(p−1))p−1<p2p），故 a<p2a<p^2a<p2。