Take the definition of Euler numbers: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!}       | r = \frac {2 e^x} {e^{2 x} + 1}       | c =  }} {{eqn | r = \paren {\frac {2 e^x} {e^{2 x} + 1 } } \paren {\frac {e^{-x} } {e^{-x} } }       | c = Multiply by $1$ }} {{eqn | r = \paren {\frac 2 {e^x + e^{-x} } }       | c =  }} {{end-eqn}} 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 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \frac {x^4} {4!} + \cdots       | c =  }} {{eqn | l = e^{-x}       | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-x}^n} {n!}       | c =  }} {{eqn | r = 1 - x + \frac {x^2} {2!} - \frac {x^3} {3!} + \frac {x^4} {4!} - \cdots       | c =  }} {{eqn | l = \paren {\frac {e^x + e^{-x} } 2}       | r = \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2 n} } {\paren {2 n}!} }       | c =  }} {{eqn | r = 1 + \frac {x^2} {2!} + \frac {x^4} {4!} + \cdots       | c = odd terms cancel in the sum. }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = 1       | r = \paren {\frac 2 {e^x + e^{-x} } } \paren {\frac {e^x + e^{-x} } 2}       | c =  }} {{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2 n} } {\paren {2 n}!} }       | c =  }} {{end-eqn}} By Product of Absolutely Convergent Series, we will let: {{begin-eqn}} {{eqn | l = a_n       | r = \frac {E_n x^n} {n!}       | c =  }} {{eqn | l = b_n       | r = \frac {x^{2 n} } {\paren {2 n}!}        | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 0}^\infty c_n       | r = \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n}       | rr= = 1       | c =  }} {{eqn | l = c_n       | r = \sum_{k \mathop = 0}^n a_k b_{n - k}       | c =  }} {{eqn | l = c_0       | r = \frac {E_0 x^0} {0!} \frac {x^0 } {0!}        | rr= = 1       | c = $c_0 = \paren {a_0} \paren {b_{0 - 0} } = \paren {a_0} \paren {b_0}$ }} {{eqn | ll= \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       | rr= = 0       | c = Subtract $1$ from both sides }} {{end-eqn}} We now have: {{begin-eqn}} {{eqn | l = c_1       | r = \frac {E_0 x^0} {0!} \frac {x^2} {2!} + \frac {E_1 x^1} {1!} \frac {x^0} {0!}       | c = $= a_0 b_1 + a_1 b_0$ }} {{eqn | l = c_2       | r = \frac {E_0 x^0} {0!} \frac {x^4} {4!} + \frac {E_1 x^1} {1!} \frac {x^2} {2!} + \frac {E_2 x^2} {2!} \frac {x^0} {0!}       | c = $= a_0 b_2 + a_1 b_1 + a_2 b_0$ }} {{eqn | l = c_3       | r = \frac {E_0 x^0} {0!} \frac {x^6} {6!} + \frac {E_1 x^1} {1!} \frac {x^4} {4!} + \frac {E_2 x^2} {2!} \frac {x^2} {2!} + \frac {E_3 x^3} {3!} \frac {x^0 } {0!}       | c = $= a_0 b_3 + a_1 b_2 + a_2 b_1 + a_3 b_0$ }} {{eqn | l = c_4       | r = \frac {E_0 x^0} {0!} \frac {x^8} {8!} + \frac {E_1 x^1} {1!} \frac {x^6} {6!} + \frac {E_2 x^2} {2!} \frac {x^4} {4!} + \frac {E_3 x^3} {3!} \frac {x^2 } {2!} + \frac {E_4 x^4} {4!} \frac {x^0} {0!}       | c = $= a_0 b_4 + a_1 b_3 + a_2 b_2 + a_3 b_1 + a_4 b_0$ }} {{eqn | o = \cdots }} {{eqn | l = c_n       | r = \frac {E_0 x^0} {0!} \frac {x^{2 n} } {\paren {2 n}!} + \frac {E_1 x^1} {1!} \frac {x^{2 n - 2} } {\paren {2 n - 2}!} + \frac {E_2 x^2} {2!} \frac {x^{2 n - 4} } {\paren {2 n - 4 }!} + \cdots + \frac {E_n x^n} {n!} \frac {x^0} {0!}       | c =  }} {{end-eqn}} Grouping terms with even exponents produces: {{begin-eqn}} {{eqn | l = \paren {\frac 1 {0! 2!} } E_0 + \paren {\frac 1 {2! 0!} } E_2       | r = 0       | c = $x^2$ term from $c_1$ and $c_2$ }} {{eqn | l = \paren {\frac 1 {0! 4!} } E_0 + \paren {\frac 1 {2! 2!} } E_2 + \paren {\frac 1 {4! 0!} } E_4       | r = 0       | c = $x^4$ term from $c_2$, $c_3$ and $c_4$ }} {{eqn | o = \cdots }} {{eqn | l = \paren {\frac 1 {0! \paren {2 n}!} } E_0 + \paren {\frac 1 {2! \paren {2 n - 2 }!} } E_2 + \paren {\frac 1 {4! \paren {2 n - 4 }!} } E_4 + \cdots + \paren {\frac 1 {\paren {2 n}! 0!} } E_{2 n}       | r = 0       | c = $x^{2 n}$ term from $c_n$, $c_{n + 1} \cdots c_{2 n}$ }} {{end-eqn}} $\forall n \in \Z_{>0}$, multiplying the coefficients of $x^{2 n}$ through by $\paren {2 n}!$ gives: :$\paren {\dfrac {\paren {2 n}! } {0! \paren {2 n}!} } E_0 + \paren {\dfrac {\paren {2 n}! } {2! \paren {2 n - 2 }!} } E_2 + \paren {\dfrac {\paren {2 n}! } {4! \paren {2 n - 4 }!} } E_4 + \cdots + \paren {\dfrac {\paren {2 n}! } {\paren {2 n}! 0!} } E_{2 n} = 0$ But those coefficients are the binomial coefficients: :$\dbinom {2 n} 0 E_0 + \dbinom {2 n} 2 E_2 + \dbinom {2 n} 4 E_4 + \dbinom {2 n} 6 E_6 + \cdots + \dbinom {2 n} {2 n} E_{2 n} = 0$ Hence the result. {{qed}}