Theorem: Pi Squared is Irrational

Pi squared ($\pi^2$) is irrational.

Proof:
{{AimForCont}} $\pi^2$ is rational.
Then $\pi^2 = \dfrac p q$ where $p$ and $q$ are integers and $q \ne 0$.
Let us define the following even function as a polynomial of degree $2 n$:
{{begin-eqn}}
{{eqn | l = \map f x
      | r = \frac {\paren {1 - x^2}^n} {n!}
      | c = 
}}
{{eqn | r = \frac {\paren {1 + x}^n \paren {1 - x}^n} {n!}
      | c = Difference of Two Squares
}}
{{end-eqn}}
From Taylor's Theorem, we have a real function on the closed interval $\closedint {-1} {1}$ which is $2 n$ times differentiable on the open interval $\openint {-1} {1}$. 
Therefore, we can express $\map f x$ as follows:
{{begin-eqn}}
{{eqn | l = \map f x
      | r = \frac {\paren {1 + x}^n \paren {1 - x}^n} {n!}
      | c = 
}}
{{eqn | n = 1
      | r = \frac 1 {0!} \map f 1 + \frac 1 {1!} \paren {x - 1} \map {f'} 1 + \frac 1 {2!} \paren {x - 1}^2 \map {f' '} 1 + \cdots + \frac 1 {\paren {2 n}!} \paren {x - 1}^{2 n} \map {f^{\paren {2 n} } } 1
      | c = 
}}
{{end-eqn}}
Let us now write the Taylor's Series Expansion of $\map g x = \dfrac {\paren {1 + x}^n} {n!}$ about $x = 1$
{{begin-eqn}}
{{eqn | l = \map g x
      | r = \frac {\paren {1 + x}^n} {n!}
      | c = 
}}
{{eqn | r = \frac 1 {0!} \map g 1 + \frac 1 {1!} \paren {x - 1} \map {g'} 1 + \frac 1 {2!} \paren {x - 1}^2 \map {g' '} 1 + \cdots + \frac 1 {\paren {n}!} \paren {x - 1}^n \map {g^{\paren n } } 1
      | c = 
}}
{{end-eqn}}
We now multiply both sides by $\paren {1 - x}^n$ to form our polynomial
{{begin-eqn}}
{{eqn | l = \map f x
      | r = \paren {1 - x}^n \map g x
      | c = 
}}
{{eqn | r = \frac {\paren {1 + x}^n \paren {1 - x}^n} {n!}
      | c = 
}}
{{eqn | r = \paren {-1}^n \frac {\paren {1 + x}^n \paren {x - 1}^n} {n!}
      | c = 
}}
{{eqn | n = 2
      | r = \paren {-1}^n \paren {\frac 1 {0!} \paren {x - 1}^n \map g 1 + \frac 1 {1!} \paren {x - 1}^{n + 1} \map {g'} 1 + \frac 1 {2!} \paren {x - 1}^{n + 2} \map {g' '} 1 + \cdots + \frac 1 {\paren {n}!} \paren {x - 1}^{2 n} \map {g^{\paren n } } 1}
      | c = 
}}
{{end-eqn}}
Comparing coefficients between equations $\paren 1$ and $\paren 2$, we observe that terms of degree less than $n$ are zero.
We therefore deduce that $\map f 1$, $\map {f'} 1$, $\map {f' '} 1$, $\cdots$, $\map {f^{\paren {n - 1} } } 1$ are all $0$.
Comparing coefficients between equations $\paren 1$ and $\paren 2$, $\forall k \in \Z: 0 \le k \le n$, we have:
{{begin-eqn}}
{{eqn | l = \frac 1 {\paren {n + k}!} \map {f^{\paren {n + k} } } 1
      | r = \paren {-1}^n  \frac 1 {\paren k!} \map {g^{\paren k } } 1
      | c = 
}}
{{eqn | r = \paren {-1}^n  \frac {2^{n - k} } {\paren {n - k}! }
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \map {f^{\paren {n + k} } } 1
      | r = \paren {-1}^n  \frac {2^{n - k} \paren {n + k}! } {\paren {n - k}! }
      | c = 
}}
{{end-eqn}}
Hence $\set {\map {f^{\paren n } } 1,\map {f^{\paren {n + 1} } } 1, \cdots, \map {f^{\paren {2 n} } } 1}$ are all integers.
Finally we observe:
{{begin-eqn}}
{{eqn | l = \map {f'} x
      | r = \frac {-2 x \paren {1 - x^2}^{n - 1} } {\paren {n - 1}!}
      | c =  Derivative of Power, Chain Rule for Derivatives
}}
{{eqn | ll = \leadsto
      | l = \map {f'} 0
      | r = 0
      | c = 
}}
{{end-eqn}}
Hence $\set {\map {f^{\paren 1 } } 0,\map {f^{\paren 2 } } 0, \cdots, \map {f^{\paren {2 n} } } 0}$ are all zero.
We establish a lemma:
=== Lemma ===
{{:Pi Squared is Irrational/Proof 2/Lemma}}{{qed|lemma}}
Now let:
:$\ds N_n = \frac \pi 4 p^n A_n = \sum_{k \mathop = 0}^n p^{n - k} \paren {-4 q}^k \map {f^{\paren {2 k} } } 1$
Since $\map f x >0$ and $\map \cos {\dfrac {\pi x} 2} > 0$ on the open interval $\openint {-1} 1$:
:$A_n > 0$
Hence $N_n$ is a positive integer.
Nevertheless, for $x \in \closedint {-1} 1$:
:$0 \le \map f x \le \dfrac 1 {n!}$
From Shape of Cosine Function and Real Cosine Function is Bounded, we have:
:$0 \le \map \cos {\dfrac {\pi x} 2} \le 1$
Hence:
:$0 < A_n < \dfrac 2 {n!}$
Therefore:
:$0 < N_n < \dfrac \pi 2 \dfrac {p^n} {n!}$
From Power over Factorial:
:$\ds \lim_{n \mathop \to \infty} \dfrac {p^n} {n!} = 0$
It follows from the Squeeze Theorem that:
:$\ds \lim_{n \mathop \to \infty} N_n = 0$
Hence for sufficiently large $n$, $N_n$ is strictly between $0$ and $1$.
This contradicts the supposition that $N_n$ is a positive integer.
Hence by Proof by Contradiction it follows that $\pi^2$ is irrational.
{{qed}}