Theorem: Bernoulli Number in terms of Euler Numbers

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
{{begin-eqn}}
{{eqn | l = B_{2 n}
      | r = \frac {2 n} {2^{2 n} \paren {2^{2 n} - 1} } \paren {\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n - 2} {2 k} E_{2 k} E_{2 n - 2 k - 2} }
      | c = 
}}
{{eqn | r = \frac {2 n} {2^{2 n} \paren {2^{2 n} - 1} } \paren {\binom {2 n - 2} 0 E_0 E_{2 n - 2} + \binom {2 n - 2} 2 E_2 E_{2 n - 4} + \binom {2 n - 2} 4 E_4 E_{2 n - 6} + \cdots + \binom {2 n - 2} {2 n - 2} E_{2 n - 2} E_0}
      | c = 
}}
{{end-eqn}}
where:
:$B_n$ denotes the $n$th Bernoulli number
:$E_n$ denotes the $n$th Euler number.

Proof:
{{begin-eqn}}
{{eqn | l = \tan x
      | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}
      | c = Power Series Expansion for Tangent Function
}}
{{eqn | ll = \leadsto
      | l = \map {\dfrac \d {\d x} } {\tan x}
      | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \, \paren {2 n - 1} x^{2 n - 2} } {\paren {2 n}!}
      | c = Derivative of Power
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \, x^{2 n - 2} } {2 n \paren {2 n - 2}!}
      | c = 
}}
{{end-eqn}}
The coefficient $a_n$ of $x^{2 n - 2}$ is:
:$\ds a_n = \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {2 n \paren {2 n - 2}!} $
From Derivative of Tangent Function, we have:
:$\map {\dfrac \d {\d x} } {\tan x} = \sec^2 x$
{{begin-eqn}}
{{eqn | l = \sec x
      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {E_{2 n} x^{2 n} } {\paren {2 n}!}
      | c = Power Series Expansion for Secant Function
}}
{{eqn | ll= \leadsto
      | l = \sec^2 x
      | r = \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {E_{2 n} x^{2 n} } {\paren {2 n}!} }^2
      | c = squaring both sides
}}
{{eqn | r = \paren {1 - \frac {E_2 x^2} {2!} + \frac {E_4 x^4} {4!} - \frac {E_6 x^6} {6!} + \cdots}  \times \paren {1 - \frac {E_2 x^2} {2!} + \frac {E_4 x^4} {4!} - \frac {E_6 x^6} {6!} + \cdots }
      | c = 
}}
{{eqn | r = 1 - x^2 \paren {\frac {E_2} {2!} + \frac {E_2} {2!} } + x^4 \paren {\frac {E_4} {4!} + \frac {E_2 \times E_2} {2! \times 2!} + \frac {E_4} {4!} } - x^6 \paren {\frac {E_6} {6!} + \frac {E_2 \times E_4} {2! \times 4!} + \frac {E_4 \times E_2} {4! \times 2!} + \frac {E_6} {6!} } + \cdots
      | c = 
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \paren {\sum_{k \mathop = 0}^{n - 1} \frac {\paren {-1}^{n - 1} } {\paren {2 k}! \paren {2 n - 2 k - 2}!} E_{2 k} E_{2 n - 2 k - 2} } x^{2 n - 2}
      | c = 
}}
{{end-eqn}}
The coefficient $a_n$ of $x^{2 n - 2}$ above is:
:$\ds a_n = \sum_{k \mathop = 0}^{n - 1} \frac 1 {\paren {2 k}! \paren {2 n - 2 k - 2}!} E_{2 k} E_{2 n - 2 k - 2}$
Equating coefficients gives us:
{{begin-eqn}}
{{eqn | l = \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {2 n \paren {2 n - 2}!}
      | r = \sum_{k \mathop = 0}^{n - 1} \frac {\paren {-1}^{n - 1} } {\paren {2 k}! \paren {2 n - 2 k - 2}!} E_{2 k} E_{2 n - 2 k - 2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = B_{2 n}
      | r = \frac {2 n} {2^{2 n} \paren {2^{2 n} - 1} } \paren {\sum_{k \mathop = 0}^{n - 1} \frac {\paren {2 n - 2}! } {\paren {2 k}! \paren {2 n - 2 k - 2}!} E_{2 k} E_{2 n - 2 k - 2} }
      | c = rearranging terms
}}
{{eqn | r = \frac {2 n} {2^{2 n} \paren {2^{2 n} - 1} } \paren {\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n - 2} {2 k } E_{2 k} E_{2 n - 2 k - 2} }
      | c = {{Defof|Binomial Coefficient}}
}}
{{end-eqn}}
{{qed}}