引理 1 将 n n n 阶行列式 D D D 上下翻转得到 D 1 D_1 D1 后， D 1 = ( − 1 ) n ( n − 1 ) 2 D D_1 = (-1)^{\frac{n(n-1)}{2}}D D1=(−1)2n(n−1)D。
计算下方行列式（
D
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D_n
Dn 为
n
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n 阶行列式）：
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D_{n+1} = |an(a−1)n⋯(a−n)nan−1(a−1)n−1⋯(a−n)n−1⋮⋮⋮aa−1⋯a−n11⋯1|
Dn+1=
anan−1⋮a1(a−1)n(a−1)n−1⋮a−11⋯⋯⋯⋯(a−n)n(a−n)n−1⋮a−n1
根据引理 1 可知，将行列式
D
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D_{n+1}
Dn+1 上下翻转后得到
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′
D'_{n+1}
Dn+1′，则有
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D'_{n+1} = (-1)^{\frac{n(n+1)}{2}}D_{n+1}
Dn+1′=(−1)2n(n+1)Dn+1。此时
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D'_{n+1} = |11⋯1aa−1⋯a−n⋮⋮⋮an−1(a−1)n−1⋯(a−n)n−1an(a−1)n⋯(a−n)n|
Dn+1′=
1a⋮an−1an1a−1⋮(a−1)n−1(a−1)n⋯⋯⋯⋯1a−n⋮(a−n)n−1(a−n)n
因为
D
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D'_{n+1}
Dn+1′ 是范德蒙德行列式，所以
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D'_{n+1} = \prod_{1 \le i < j \le n+1} (a-i) - (a-j) = \prod_{1 \le i < j \le n+1} (j-i)
Dn+1′=1≤i<j≤n+1∏(a−i)−(a−j)=1≤i<j≤n+1∏(j−i)