5

D_{n} = | \begin{array}{cccc} 1 & 1 & \dots & 1 \\ 1 & (\binom{2}{1}) & \dots & (\binom{n}{1}) \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 1 & (\binom{n}{n - 1}) & \dots & (\binom{2 n - 2}{n - 1}) \end{array} |

\begin{matrix} = | \begin{array}{cccc} 1 & 1 & \dots & 1 \\ 0 & (\binom{1}{1}) & \dots & (\binom{n - 1}{1}) \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & (\binom{n - 1}{n - 1}) & \dots & (\binom{2 n - 3}{n - 1}) \end{array} | & (o d k a \overset{ˇ}{z} d \overset{ˊ}{e} h o \overset{ˇ}{r} \overset{ˊ}{a} d k u k r o m \overset{ˇ}{e} p r v n \overset{ˊ}{ı} h o o d e \overset{ˇ}{c} t e m e p \overset{ˇ}{r} e d c h o z \overset{ˊ}{ı}) \end{matrix}

\begin{matrix} = | \begin{array}{cccc} (\binom{1}{1}) & (\binom{2}{1}) & \dots & (\binom{n - 1}{1}) \\ (\binom{2}{2}) & (\binom{3}{2}) & \dots & (\binom{n}{2}) \\ ⋮ & ⋮ & ⋱ & ⋮ \\ (\binom{n - 1}{n - 1}) & (\binom{n}{n - 1}) & \dots & (\binom{2 n - 3}{n - 1}) \end{array} | & (r o z v i n e m e p o d l e p r v n \overset{ˊ}{ı} h o s l o u p c e) \end{matrix}

\begin{matrix} = | \begin{array}{cccc} 1 & 1 & \dots & 1 \\ 1 & (\binom{2}{1}) & \dots & (\binom{n - 1}{1}) \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 1 & (\binom{n - 1}{n - 2}) & \dots & (\binom{2 n - 4}{n - 2}) \end{array} | & (o d k a \overset{ˇ}{z} d \overset{ˊ}{e} h o s l o u p c e k r o m \overset{ˇ}{e} p r v n \overset{ˊ}{ı} h o o d e \overset{ˇ}{c} t e m e p \overset{ˇ}{r} e d c h o z \overset{ˊ}{ı}) \end{matrix}

= D_{n - 1}

Ze základního případu $D_{1} = | \begin{array}{c} 1 \end{array} | = 1$ plyne $(\forall n \in ℕ^{+}) (D_{n} = 1)$ .