Take the definition of Euler numbers: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!}       | r = \frac {2e^x} {e^{2x} + 1}       | c =  }} {{eqn | r = \paren {\frac {2e^x } {e^{2x} + 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^{2n } } {\paren {2n }!} }       | 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^{2n } } {\paren {2n }!} }       | 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^{2n} } {\paren {2n }!}        | 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 {E_0 x^0} {0!} \frac {x^{0} } {0!}        | 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. }} {{end-eqn}} $\forall n \in \Z_{\gt 0}$, term by term $c_n$ is equal to: {{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!}       | rrr = = \frac {x^2 } {2! } E_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!}       | rrr = = \frac {x^4 } {4! } E_0 + \frac {x^2 } {2! } E_2       | 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!}       | rrr = = \frac {x^6 } {6! } E_0 + \frac {x^4 } {2! 2! } E_2       | 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!}       | rrr = = \frac {x^8 } {0! 8! } E_0 + \frac {x^6 } {2! 4! } E_2 + \frac {5 x^4 } {4! 0! } E_4       | c = $= a_0 b_4 + a_1 b_3 + a_2 b_2 + a_3 b_1 + a_4 b_0$ }} {{eqn | l = \cdots       | r = \cdots }} {{eqn | l = c_n       | r = \frac {E_0 x^0} {0!} \frac {x^{2n} } {\paren {2n }!} + \frac {E_1 x^1} {1!} \frac {x^{2n - 2} } {\paren {2n - 2 }!} + \frac {E_2 x^2} {2!} \frac {x^{2n - 4} } {\paren {2n - 4 }!} + \cdots + \frac {E_n x^n} {n!} \frac {x^0 } {0!}       | c =  }} {{end-eqn}} Grouping like even terms 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 }} {{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 }} {{end-eqn}} Multiplying $c_n$ through by $n!$ gives: {{begin-eqn}} {{eqn | l = n! c_n       | r = \frac {E_0 x^0} {0!} \frac {n! x^n } {n!} + \frac {E_1 x^1} {1!} \frac {n! x^{n-1} } {\paren {n - 1 }!} + \cdots + \frac {E_n x^n} {n!} \frac {n! x^{0} } {0!}       | rrr = = 0       | c =  }} {{eqn | r = x^n \paren {\frac {n! } {0! n!} E_0 + \frac {n! } {1! \paren {n - 1 }!} E_1 + \cdots + \frac {n! } {n! 0!} E_n }       | rrr = = 0       | c = factoring out $x^n$ }} {{end-eqn}} But those coefficients are the binomial coefficients: {{begin-eqn}} {{eqn | l = n! c_n       | r = \dbinom n 0 E_0 + \dbinom n 1 E_1 + \dbinom n 2 E_2 + \cdots + \dbinom n n E_n       | rrr = = 0       | c =  }} {{end-eqn}} Hence the result. {{qed}} {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n} \dbinom {2 n} {2 k} E_{2 n - 2 k}       | r = \binom {2 n} 0 E_{2 n} + \binom {2 n} 2 E_{2 n - 2} + \binom {2 n} 4 E_{2 n - 4} + \binom {2 n} 6 E_{2 n - 6} + \cdots + 1       | rrr = = 0       | c =  }} {{end-eqn}}