For $z \in \C$ and $n \in \N_{> 0}$, let: :$\map {f_n} z = \dfrac 1 2 \paren {\paren {1 + \dfrac z n}^n - \paren {1 - \dfrac z n}^n}$ Then $\map {f_n} z = 0$ {{iff}}: {{begin-eqn}} {{eqn | l = \paren {1 + \dfrac z n}^n       | r = \paren {1 - \dfrac z n}^n       | c =  }} {{eqn | ll= \leadstoandfrom       | l = 1 + \frac z n        | r = \paren {1 - \dfrac z n} e^{2 \pi i \frac k n}       | c = for $k \in \Z$ }} {{eqn | ll= \leadstoandfrom       | l = z       | r = n \frac {e^{2 \pi i \frac k n} - 1} {e^{2 \pi i \frac k n} + 1}       | c =  }} {{eqn | r = n i \map \tan {\frac {k \pi} n}       | c = Tangent Exponential Formulation }} {{end-eqn}} Let $n = 2 m + 1$. Then the roots of $\map {f_{2 m + 1} } z$ are: :$\paren {2 m + 1} i \map \tan {\dfrac {k \pi} {2 m + 1} }$  for $- m \le k \le m$. Observe that $\map {f_{2 m + 1} } z$ is a polynomial of degree $2 m + 1$.  Then for some constant $C$, we have: {{begin-eqn}} {{eqn | l = \map {f_{2 m + 1} } z       | r = C z \prod_{\substack {k \mathop = - m \\ k \mathop \ne 0} }^m \paren {1 - \frac z {\paren {2 m + 1} i \map \tan {k \pi / \paren {2 m + 1} } } }       | c = Polynomial Factor Theorem }} {{eqn | r = C z \prod_{k \mathop = 1}^m \paren {1 + \frac {z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } }       | c = Tangent Function is Odd }} {{end-eqn}} It can be seen from the Binomial Theorem that the coefficient of $z$ in $\map {f_{2 m + 1} } z$ is $1$. Hence $C = 1$, and we obtain: :$\displaystyle \map {f_{2 m + 1} } z = z \prod_{k \mathop = 1}^m \paren {1 + \frac {z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } }$ First we consider $z = x$ where $x$ is a non-negative real number. Let $l < m$. Then: :$\displaystyle x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } \le \map {f_{2 m + 1} } x$ Taking the limit as $m \to \infty$ we have: {{begin-eqn}} {{eqn | l = \lim_{m \mathop \to \infty} x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} \paren {\frac {k \pi / \paren {2 m + 1} } {\map \tan {k \pi / \paren {2 m + 1} } } }^2}       | o = \le        | r = \frac 1 2 \paren {e^x - e^{-x} }       | c = {{Defof|Exponential Function/Complex|Exponential Function|subdef = Limit of Sequence}} }} {{eqn | ll= \leadsto       | l = x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} }       | o = \le       | r = \sinh x       | c = Limit of Tan X over X and {{Defof|Hyperbolic Sine}} }} {{end-eqn}} By Tangent Inequality, we have: :$\map \tan {\dfrac {k \pi} {2 m + 1} } \ge \dfrac {k \pi} {2 m + 1}$ hence: :$\displaystyle \map {f_{2 l + 1} } x \le x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} } \le \sinh x$ Taking the limit as $l \to \infty$ we have by Squeeze Theorem: :$(1): \quad \displaystyle x \prod_{k \mathop = 1}^\infty \paren {1 + \frac {x^2} {k^2 \pi^2} } = \sinh x$ Now let $1 < l < m$. By Complex Modulus of Product of Complex Numbers and the Triangle Inequality, we can deduce: {{begin-eqn}} {{eqn | r = \cmod {\map {f_{2 m + 1} } z - z \prod_{k \mathop = 1}^l \paren {1 + \frac {z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } }       | o =  }} {{eqn | r = \cmod z \cmod {\prod_{k \mathop = 1}^l \paren {1 + \frac {z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } } \cdot \cmod {\prod_{k \mathop = l + 1}^m \paren {1 + \frac {z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } - 1}       | c =  }} {{eqn | o = \le        | r = \cmod z \paren {\prod_{k \mathop = 1}^l \paren {1 + \frac {\cmod z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } } \cdot \paren {\prod_{k \mathop = l + 1}^m \paren {1 + \frac {\cmod z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } - 1}       | c =  }} {{eqn | r = \map {f_{2 m + 1} } {\cmod z} - \cmod z \prod_{k \mathop = 1}^l \paren {1 + \frac {\cmod z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } }       | c =  }} {{end-eqn}} Taking the limit as $m \to \infty$ we have: :$\displaystyle \cmod {\sinh z - z \prod_{k \mathop = 1}^l \paren {1 + \frac {z^2} {k^2 \pi^2} } } \le \sinh \cmod z - \cmod z \prod_{k \mathop = 1}^l \paren {1 + \frac {\cmod z^2} {k^2 \pi^2} }$ Now take the limit as $l \to \infty$. By $(1)$ and Squeeze Theorem, we have: :$\displaystyle \sinh z = z \prod_{k \mathop = 1}^\infty \paren {1 + \frac {z^2} {k^2 \pi^2} }$ Finally, substituting $z \mapsto i z$, we obtain: :$\displaystyle \sin z = z \prod_{k \mathop = 1}^\infty \paren {1 - \frac {z^2} {k^2 \pi^2} }$ {{qed}}