Take the definition of Bernoulli numbers: :$\displaystyle \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$ From the definition of the exponential function: {{begin-eqn}} {{eqn | l = e^x       | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}       | c =  }} {{eqn | r = 1 + \sum_{n \mathop = 1}^\infty \frac {x^n} {n!}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {e^x - 1} x       | r = \sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {n!}       | c =  }} {{eqn | r = 1 + \frac x {2!} + \frac {x^2} {3!} + \cdots       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = 1       | r = \paren {\frac x {e^x - 1} } \paren {\frac {e^x - 1} x}       | c =  }} {{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} } \paren {\sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {n!} }       | c =  }} {{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^{n} } {\paren{n + 1 }!} }       | c = Both series start at zero }} {{end-eqn}} By Product of Absolutely Convergent Series, we will let: {{begin-eqn}} {{eqn | l = a_n       | r = \frac {B_n x^n} {n!}       | c =  }} {{eqn | l = b_n       | r = \frac {x^{n} } {\paren {n + 1 }!}        | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 0}^\infty c_n       | r = \paren { \displaystyle \sum_{n \mathop = 0}^\infty a_n } \paren {\displaystyle \sum_{n \mathop = 0}^\infty b_n }       | rrr = =1       | c =  }} {{eqn | l = c_n       | r = \sum_{k \mathop = 0}^n a_k b_{n - k}       | c =  }} {{eqn | l = c_0       | r = \frac {B_0 x^0} {0!} \frac {x^{0} } {\paren {0 + 1 }!}        | rrr = = 1       | c = $c_0 = \paren {a_0 } \paren {b_{0 - 0 } } = \paren {a_0 } \paren {b_0 }$ }} {{eqn | lll = \leadsto       | l = \sum_{n \mathop = 1}^\infty c_n       | r = \paren { \displaystyle \sum_{n \mathop = 0}^\infty a_n } \paren {\displaystyle \sum_{n \mathop = 0}^\infty b_n } - a_0 b_0       | rrr = =0       | c = Subtract 1 from both sides of the equation. }} {{eqn | r = c_1x + c_2x^2 + c_3x^3 + \cdots       | rrr = =0 }} {{end-eqn}} $\leadsto \forall n \in \Z_{\gt 0}$, $c_n = 0$. {{begin-eqn}} {{eqn | l = c_1       | r = \frac {B_0 x^0} {0!} \frac {x^{1} } {\paren {1 + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{0} } {\paren {0 + 1 }!}       | rrr = = 0       | c = $= a_0 b_1 + a_1 b_0$ }} {{eqn | l = c_2       | r = \frac {B_0 x^0} {0!} \frac {x^{2} } {\paren {2 + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{1} } {\paren {1 + 1 }!} + \frac {B_2 x^2} {2!} \frac {x^{0} } {\paren {0 + 1 }!}       | rrr = = 0       | c = $= a_0 b_2 + a_1 b_1 + a_2 b_0$ }} {{eqn | l = \cdots       | r = \cdots       | rrr = = 0 }} {{eqn | l = c_n       | r = \frac {B_0 x^0} {0!} \frac {x^{n} } {\paren {n + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{n-1} } {\paren {n - 1 + 1 }!} + \cdots + \frac {B_n x^n} {n!} \frac {x^{0} } {\paren {0 + 1 }!}       | rrr = = 0       | c = $= a_0 b_n + a_1 b_{n - 1 } + a_2 b_{n - 2 } + \cdots + a_n b_0$ }} {{end-eqn}} Multiplying $c_n$ through by $\paren {n + 1 }!$ gives: {{begin-eqn}} {{eqn | l = \paren {n + 1 }! c_n       | r = \frac {B_0 x^0} {0!} \frac {\paren {n + 1 }! x^n } {\paren {n + 1 }!} + \frac {B_1 x^1} {1!} \frac {\paren {n + 1 }! x^{n-1} } {\paren {n - 1 + 1 }!} + \cdots + \frac {B_n x^n} {n!} \frac {\paren {n + 1 }! x^{0} } {\paren {0 + 1 }!}       | rrr = = 0       | c =  }} {{eqn | r = x^n \paren {\frac {\paren {n + 1 }! } {0! \paren {n + 1 }!} B_0 + \frac {\paren {n + 1 }! } {1! \paren {n - 1 + 1 }!} B_1 + \cdots + \frac {\paren {n + 1 }! } {n! \paren {0 + 1 }!} B_n }       | rrr = = 0       | c = factoring out $x^n$ }} {{end-eqn}} But those coefficients are the binomial coefficients: {{begin-eqn}} {{eqn | l = \paren {n + 1 }! c_n       | r = \dbinom {n + 1 } 0 B_0 + \dbinom {n + 1 } 1 B_1 + \dbinom {n + 1 } 2 B_2 + \cdots + \dbinom {n + 1 } n B_n       | rrr = = 0       | c =  }} {{eqn | l = n! c_{n-1 }       | r = \dbinom n 0 B_0 + \dbinom n 1 B_1 + \dbinom n 2 B_2 + \cdots + \dbinom n {n - 1} B_{n - 1}       | rrr = = 0       | c =  }} {{end-eqn}} Hence the result. {{qed}}