These are the proofs that I have seen:

The proof using Fourier series: Reference: Stein Shakarchi "Fourier Analysis, the introduction" p 97 Exercise 4 The key ingredient of this proof is the following identity $$\sum_{n=1}^{\infty} \frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan{\alpha\pi}}$$ which can be proved by using Fourier series of $\cos(\alpha x)$. This allows an expression of $\zeta(2n)$ by Bernoulli numbers.

The proof using the functional equation of $\zeta(s)$: Reference: E.Titchmarsh "The Theory of the Riemann zeta function" p18 (2.4 Second method)

Chapter 2 of this book is entirely devoted to the proofs of the functional equation $$\pi^{-s/2}\Gamma(s)\zeta(s)=\pi^{-(1-s)/2}\Gamma(1-s)\zeta(1-s).$$ Section 2.4 is one of the proofs, it uses te residue theorem of complex analysis, and derives the formula $$\zeta(1-2m)=\frac{(-1)^mB_{2m}}{2m}$$ for $m=1,2,\cdots$ The formula for $\zeta(2n)$ is now followd by the functional equation.

The result is: $$2\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{(2n)!}B_{2n}$$ where $$\frac{z}{e^z-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}z^n.$$

These are the proofs that I have seen:

The proof using Fourier series: Reference: Stein Shakarchi "Fourier Analysis, the introduction" p 97 Exercise 4 The key ingredient of this proof is the following identity $$\sum_{n=1}^{\infty} \frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan{\alpha\pi}}$$ which can be proved by using Fourier series of $\cos(\alpha x)$. This allows an expression of $\zeta(2n)$ by Bernoulli numbers.

The proof using the functional equation of $\zeta(s)$: Reference: E.Titchmarsh "The Theory of the Riemann zeta function" p18 (2.4 Second method)

Chapter 2 of this book is entirely devoted to the proofs of the functional equation $$\pi^{-s/2}\Gamma(s/2)\zeta(s)=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s).$$ Section 2.4 is one of the proofs, it uses te residue theorem of complex analysis, and derives the formula $$\zeta(1-2m)=\frac{(-1)^mB_{2m}}{2m}$$ for $m=1,2,\cdots$ The formula for $\zeta(2n)$ is now followd by the functional equation.

The result is: $$2\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{(2n)!}B_{2n}$$ where $$\frac{z}{e^z-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}z^n.$$

These are the proofs that I have seen:

The proof using Fourier series: Reference: Stein Shakarchi "Fourier Analysis, the introduction" p 97 Exercise 4 The key ingredient of this proof is the following identity $$\sum_{n=1}^{\infty} \frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan{\alpha\pi}}$$ which can be proved by using Fourier series of $\cos(\alpha\pi)$. This allows an expression of $\zeta(2n)$ by Bernoulli numbers.

The proof using the functional equation of $\zeta(s)$: Reference: E.Titchmarsh "The Theory of the Riemann zeta function" p18 (2.4 Second method)

Chapter 2 of this book is entirely devoted to the proofs of the functional equation $$\pi^{-s/2}\Gamma(s)\zeta(s)=\pi^{-(1-s)/2}\Gamma(1-s)\zeta(1-s).$$ Section 2.4 is one of the proofs, it uses te residue theorem of complex analysis, and derives the formula $$\zeta(1-2m)=\frac{(-1)^mB_{2m}}{2m}$$ for $m=1,2,\cdots$ The formula for $\zeta(2n)$ is now followd by the functional equation.

The result is: $$2\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{(2n)!}B_{2n}$$ where $$\frac{z}{e^z-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}z^n.$$

These are the proofs that I have seen:

The proof using Fourier series: Reference: Stein Shakarchi "Fourier Analysis, the introduction" p 97 Exercise 4 The key ingredient of this proof is the following identity $$\sum_{n=1}^{\infty} \frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan{\alpha\pi}}$$ which can be proved by using Fourier series of $\cos(\alpha x)$. This allows an expression of $\zeta(2n)$ by Bernoulli numbers.

The proof using the functional equation of $\zeta(s)$: Reference: E.Titchmarsh "The Theory of the Riemann zeta function" p18 (2.4 Second method)

Chapter 2 of this book is entirely devoted to the proofs of the functional equation $$\pi^{-s/2}\Gamma(s)\zeta(s)=\pi^{-(1-s)/2}\Gamma(1-s)\zeta(1-s).$$ Section 2.4 is one of the proofs, it uses te residue theorem of complex analysis, and derives the formula $$\zeta(1-2m)=\frac{(-1)^mB_{2m}}{2m}$$ for $m=1,2,\cdots$ The formula for $\zeta(2n)$ is now followd by the functional equation.

The result is: $$2\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{(2n)!}B_{2n}$$ where $$\frac{z}{e^z-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}z^n.$$