( 1 + 1 1 ) ( 1 + 1 2 ) ( 1 + 1 3 + ⋯ + ( 1 + 1 n ) = n + 1 {\displaystyle (1+{\frac {1}{1}})(1+{\frac {1}{2}})(1+{\frac {1}{3}}+\cdots +(1+{\frac {1}{n}})=n+1}
L : 1 + 1 n = 1 + 1 1 = 2 R : n + 1 = 1 + 1 = 2 2 = 2 {\displaystyle {\begin{aligned}&L:1+{\frac {1}{n}}=1+{\frac {1}{1}}=2\\&R:n+1=1+1=2\\&2=2\end{aligned}}}
( 1 + 1 1 ) ( 1 + 1 2 ) ( 1 + 1 3 + ⋯ + ( 1 + 1 k ) = k + 1 {\displaystyle (1+{\frac {1}{1}})(1+{\frac {1}{2}})(1+{\frac {1}{3}}+\cdots +(1+{\frac {1}{k}})=k+1}
( 1 + 1 1 ) ( 1 + 1 2 ) ( 1 + 1 3 + ⋯ + ( 1 + 1 k ) ⏟ = k + 1 ∗ ( 1 + 1 k + 1 ) = k + 2 ( k + 1 ) ∗ ( 1 + 1 k + 1 ) = k + 2 k + 1 + k + 1 k + 1 = k + 2 0 = 0 {\displaystyle {\begin{aligned}&\underbrace {(1+{\frac {1}{1}})(1+{\frac {1}{2}})(1+{\frac {1}{3}}+\cdots +(1+{\frac {1}{k}})} _{=k+1}*(1+{\frac {1}{k+1}})=k+2\\&(k+1)*(1+{\frac {1}{k+1}})=k+2\\&k+1+{\frac {k+1}{k+1}}=k+2\\&0=0\\\end{aligned}}}
הטענה נכונה עבור כל n טבעי, ע"פ שלושת שלבי האינדוקציה.