Theorem: Power Series Expansion for Exponential of Cosine of x

{{begin-eqn}}
{{eqn | l = e^{\cos x}
      | r = e \paren {e^{\cos x - 1} }
      | c = 
}}
{{eqn | r = e \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {-1}^m \map {P_2} {2 m} } {2 m!} x^{2 m} }
      | c = where $\map {P_2} {2 m}$ is the partition of the set of size $2 m$ into even blocks
}}
{{eqn | r = e \paren {1 - \dfrac {x^2} 2 + \dfrac {x^4} 6 - \dfrac {31 x^6} {720} + \cdots}
      | c = 
}}
{{end-eqn}}
for all $x \in \R$.

Proof:
A result in combinatorics known as the Exponential formula states that if 
:$\ds \map f x = \sum_{n \mathop = 1}^\infty \frac {c_n} {n!} x^n $
then:
:$\ds e^{\map f x} = \sum_{k \mathop = 0}^\infty \frac {\map {B_k} {c_1, c_2, \ldots, c_k} } {k!} x^k$
where $\map {B_k} {c_1, c_2, \ldots, c_k}$ is the $k$th complete Bell Polynomial.
From the Power Series Expansion for Cosine Function, we get:
:$\ds \cos x - 1 = \sum_{p \mathop = 1}^\infty \frac {\paren {-1}^p} {2 p!} x^{2 p}$
Using this we may plug the sequence $\tuple {0, -1, 0, 1, 0, -1}$ into the first $6$ complete Bell Polynomials.
An arbitrary even Complete Bell Polynomial, will take the form:
:$\ds \map {B_n} {x_1, x_2, x_3, \ldots, x_n} = \sum \frac {n!} {k_1! k_2! \dotsm k_n!} \paren {\frac {x_1} {1!} }^{c_1} \paren {\frac {x_2} {2!} }^{c_2} \dotsm \paren {\frac {x_n} {n!} }^{c_n}$
where the sum is taken over all n-tuples $\tuple {c_1, \ldots, c_n}$ such that $c_1 + 2 c_2 + \dotsb + n c_n = n$
In other words, it is taken over every integer partition of $n$.
Here we have it that
:$x_{4 n + 0} = 1$
:$x_{4 n + 1} = 0$
:$x_{4 n + 2} = -1$
:$x_{4 n + 3} = 0$
Because there is no way to form a partition of an odd number without using an odd number, all the summands in the odd Complete Bell polynomials contain a $0$, thus they equal $0$.
Explicitly calculating the first $6$ even complete Bell Polynomials we get:
:$\map {B_0} 0 = 1$
:$\map {B_2} {0, -1} = 0^2 - 1 = -1$
:$\map {B_4} {0, -1, 0, 1} = 0^4 + 6 \paren 0^2 \paren {-1} + 4 \paren 0 \paren 0 + 3 \paren {-1}^2 + 1 = 4$
:$\map {B_6} {0, -1, 0, 1, 0, -1} = 0^6 + 6 \paren {-1} \paren 0 + 15 \paren {-1} (1) + 10 \paren 0^2 + 15 \paren 0^2 (1) + 15 \paren {-1}^3 + 60 \paren 0 \paren {-1} \paren 0 + 20 \paren 0^3 \paren 0 + 45 \paren 0^2 \paren {-1}^2 + 15 \paren 0^4 \paren {-1} + \paren {-1} = -31$
Note that in all the summands for the above polynomials, they are either $0$ or of the same sign.
This is provable to always be the case, as if $n \equiv 2 \pmod 4$, then an odd number of integers congruent to $2 \pmod 4$ must be chosen for every partition.
Thus all summands will be a negative.
A similar argument holds for the case $n \equiv 0 \pmod 4$.
Thus for our given values of $x_k$, we are only summing over the partitions of $2 m$ such that all summands in the partition are even, and because all the summands in the polynomial are the same sign we have:
:$\map {B_{2 m} } {0, -1, 0, \ldots, \pm 1} = \paren {-1}^m \map {B_{2 m} } {0, 1, 0, 1, \ldots, 1}$
Thus from the definition of Bell Polynomials as a sum over all incomplete Bell polynomials we have:
:$\map {B_{2 m} } {0, -1, 0, \ldots , \pm x_{2 m} } = \paren {-1}^m \map {P_2} {2 m}$
where $\map {P_2} {2 m}$ is all partitions of a set of size $2 m$ into even blocks.
Thus we have:
:$\ds e^{\cos x} = e \paren {e^{\cos x - 1} } = e \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {-1}^m P_2 \paren {2 m} } {2m!} x^{2 m} }$
where $P_2 \paren {2 m}$ is the partition of the set of size $2 m$ into even blocks.
{{qed}}