Obfuscated proofs I am, just for fun, looking for long and complicated proofs for statements which can be proven rather easily and much faster. The proof itself still has to be correct however.
While the proof should be obfuscated, all parts should have some relevance. So do not prove Fermat‘s last theorem and end with "ah, by the way: 1+1=2, so the statement follows.
It is also boring to obfuscate simple arithmetic; one can prove "1+1=2" in 100 pages only using addition, subtraction, multiplication and division – but that is not fun.
I rather look for some very interesting obfuscation of a proof. Maybe a statement of elementary number theory can be proven in a "nice" complicated way. Or maybe one can use functional analysis to prove basic analysis stuff etc.

Here is one (not that good) example: Theorem: For $a, b \in \mathbb{R}$ it holds that $(a-b)^2 = a^2 - 2ab + b^2$.
Proof: Let $f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto x^2$. As $f$ is analytical the Taylor expansion of $f$ converges. Therefore
$$
\begin{align*}
    x^2
 &= f(x) \\
 &= Tf(x,b) \\
 &= \sum_{n=0}^\infty \frac{f^{(n)}(b)}{n!} (x-b)^n \\
 &= \frac{b^2}{0!} + \frac{2b}{1!} (x-b) + \frac{2}{2!} (x-b)^2 + \sum_{n=3}^\infty \frac{0}{n!} (x-b)^n\\
 &= b^2 + 2bx - 2b^2 + (x-b)^2 \\
 &= -b^2 + 2bx + (x-b)^2,
\end{align*}
$$
ie.
$$
    x^2 + b^2 - 2bx = (x-b)^2
$$
and for $x = a$ the theorem follows.
 A: Maybe Wilson's theorem. For any odd prime $p$, $(p-1)!=-1$ mod $p$.
Easy proof : Among $2,..., p-2$ the numbers can be gathered by pairs of inverse mod $p$ so that : 
$$2\times...\times p-2=1\text{ mod }p $$
Hence : $(p-1)!=-1$ mod $p$.
Slighty more complicated proof : Take $G:=\mathfrak{S}_p$ the symmetric group of $p$ elements. Set $n_p$ the number of $p$-Sylows of $G$.
We know that its $p$-Sylows are cyclic groups of order $p$. By a standard argument we know that there are $(p-1)!$ $p$-cycles. A $p$-cycle will have exactly $p-1$ $p$-cycles in the $p$-Sylow it generates so that $n_p=\frac{(p-1)!}{p-1}=(p-2)!$.
Now we apply second Sylow's theorem to get $n_p=1$ mod $p$. Finally $(p-1)!=-1$ mod $p$ by multiplying both sides by $p-1$.
A: Every function from $\mathbb{N}$ to $\mathbb{R}$ is continuous, where the topologies are the canonical.
Proof 1: [Short]
$\mathbb{N}$ is a discrete space. $\blacksquare$
Proof 2:
Take a function $f: \mathbb{N} \rightarrow \mathbb{R}$. Now, consider the following function $g:\mathbb{R} \rightarrow \mathbb{R}$:
$g(x)=\left(f(\lfloor x \rfloor +1)-f(\lfloor x \rfloor) \right)(x-\lfloor x \rfloor)+f(\lfloor x \rfloor)$
In each interval $[n,n+1]$, where $n \in \mathbb{N}$, we have that the function $g$ is a linear function, hence continuous. Hence, by the pasting lemma, $g$ is a continuous function. Now, it is easy to see that $g|_{\mathbb{N}}=f$. Since the restriction of a continuous function is continuous, we have our result. $\blacksquare$
Also, see https://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts
and more specifically:
https://mathoverflow.net/a/44742/48745 (which is a gem)
A: Proof: $0a = 0$
$$(a+0)^2 = a^2 = a^2 + 2(0) + 0^2$$
$$a^2 = a^2 + 2(0) + 0^2$$ $$0 = 2(0) + 0^2$$ $$2(0) = (1+1)0 = 1(0) + 1(0) = 0 + 0 = 0$$ $$\dfrac{0^2}{2} >= \sqrt{0^2} = 0$$ $$0^2 >= 2(0)$$
We have proven that $$2(0) = 0$$
Therefore, $$(a+0)^2 = a^2 = a^2 + 2(0) + 0^2 = a^2 + 0$$ $$a^2 = a^2 + 0$$ $$0 = 0$$
Q.E.D.
