Theorem: Power Series Expansion for Exponential of Tangent of x

:$e^{\tan x} = 1 + x + \dfrac {x^2} 2 + \dfrac {x^3} 2 + \dfrac {3 x^4} 8 + \cdots$
for all $x \in \R$ such that $\size x < \dfrac \pi 2$.

Proof:
Let $\map f x = e^{\tan x}$.
Then:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} \map f x
      | r = \sec^2 x \, e^{\tan x}
      | c = Chain Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \frac {\d^2} {\d x^2} \map f x
      | r = \sec^2 x \frac \d {\d x} e^{\tan x} + e^{\tan x} \frac \d {\d x} \sec^2 x
      | c = Product Rule for Derivatives
}}
{{eqn | r = \sec^4 x \, e^{\tan x} + e^{\tan x} 2 \sec^2 x \tan x
      | c = 
}}
{{eqn | r = \paren {\sec^2 x + 2 \tan x} \sec^2 x \, e^{\tan x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d^3} {\d x^3} \map f x
      | r = \paren {\sec^2 x + 2 \tan x} \frac \d {\d x} \sec^2 x \, e^{\tan x} + \sec^2 x \, e^{\tan x} \frac \d {\d x} \paren {\sec^2 x + 2 \tan x}
      | c = Product Rule for Derivatives
}}
{{eqn | r = \paren {\sec^2 x + 2 \tan x} \paren {\sec^2 x + 2 \tan x} \sec^2 x \, e^{\tan x} + \sec^2 x \, e^{\tan x} \paren {2 \sec^2 x \tan x + 2 \sec^2 x}
      | c = 
}}
{{eqn | r = \paren {\paren {\sec^2 x + 2 \tan x}^2 + 2 \sec^2 x \tan x + 2 \sec^2 x} \sec^2 x \, e^{\tan x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d^4} {\d x^4} \map f x
      | r = \paren {\paren {\sec^2 x + 2 \tan x}^2 + 2 \sec^2 x \tan x + 2 \sec^2 x} \frac \d {\d x} \sec^2 x \, e^{\tan x}
      | c = Product Rule for Derivatives
}}
{{eqn | o = 
      | ro= +
      | r = \sec^2 x \, e^{\tan x} \frac \d {\d x} \paren {\paren {\sec^2 x + 2 \tan x}^2 + 2 \sec^2 x \tan x + 2 \sec^2 x}
      | c = 
}}
{{eqn | r = \paren {\paren {\sec^2 x + 2 \tan x}^2 + 2 \sec^2 x \tan x + 2 \sec^2 x} \paren {\sec^2 x + 2 \tan x} \sec^2 x \, e^{\tan x}
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \sec^2 x \, e^{\tan x} \paren {2 \paren {\sec^2 x + 2 \tan x} \paren {2 \sec^2 x \tan x + 2 \sec^2 x} + 4 \sec^2 x \tan^2 x + 2 \sec^4 x + 4 \sec^2 x \tan x}
      | c = 
}}
{{end-eqn}}
By definition of Taylor series:
:$\ds \map f x \sim \sum_{n \mathop = 0}^\infty \frac {\paren {x - \xi}^n} {n!} \map {f^{\paren n} } \xi$
This is to be expanded about $\xi = 0$.
Note that $\tan 0 = 0$ and $\sec 0 = 1$.
Thus:
{{begin-eqn}}
{{eqn | l = \map f 0
      | r = e^0
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | l = \valueat {\frac \d {\d x} \map f x} {x \mathop = 0}
      | r = \sec^2 0 \, e^{\tan 0}
      | c = 
}}
{{eqn | r = 1 \times e^0
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | l = \valueat {\frac {\d^2} {\d x^2} \map f x} {x \mathop = 0}
      | r = \paren {\sec^2 0 + 2 \tan 0} \sec^2 0 \, e^{\tan 0}
      | c = 
}}
{{eqn | r = \paren {1 + 0} 1 \times e^0
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | l = \valueat {\frac {\d^3} {\d x^3} \map f x} {x \mathop = 0}
      | r = \paren {\paren {\sec^2 0 + 2 \tan 0}^2 + 2 \sec^2 0 \tan 0 + 2 \sec^2 0} \sec^2 0 \, e^{\tan 0}
      | c = 
}}
{{eqn | r = \paren {\paren {1 + 0}^2 + 2 \times 1 \times 0 + 2} 1 \times e^0
      | c = 
}}
{{eqn | r = 3
      | c = 
}}
{{eqn | l = \valueat {\frac {\d^4} {\d x^4} \map f x} {x \mathop = 0}
      | r = \paren {\paren {\sec^2 0 + 2 \tan 0}^2 + 2 \sec^2 0 \tan 0 + 2 \sec^2 0} \paren {\sec^2 0 + 2 \tan 0} \sec^2 0 \, e^{\tan 0}
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \sec^2 0 \, e^{\tan 0} \paren {2 \paren {\sec^2 0 + 2 \tan 0} \paren {2 \sec^2 0 \tan 0 + 2 \sec^2 0} + 4 \sec^2 0 \tan^2 0 + 2 \sec^4 0 + 4 \sec^2 0 \tan 0}
      | c = 
}}
{{eqn | r = \paren {\paren {1 + 0}^2 + 0 + 2} \paren {1 + 0} + \paren {2 \paren {1 + 0} \paren {0 + 2} + 0 + 2 + 0}
      | c = 
}}
{{eqn | r = 9
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \map f x
      | o = \sim
      | r = \sum_{n \mathop = 0}^\infty \frac {\paren {x - \xi}^n} {n!} \map {f^{\paren n} } \xi
      | c = 
}}
{{eqn | r = \paren {1 \times 1} + \paren {x \times 1} + \paren {\dfrac {x^2} {2!} \times 1} + \paren {\dfrac {x^3} {3!} \times 3} + \paren {\dfrac {x^4} {4!} \times 9} + \cdots
      | c = 
}}
{{eqn | r = 1 + x + \dfrac {x^2} 2 + \dfrac {x^3} 2 + \dfrac {3 x^4} 8 + \cdots
      | c = 
}}
{{end-eqn}}
{{qed}}