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Question: Why are the roots of $\frac {\sin x}x$ equal to $\frac x{k\pi}$ for $k\in\mathbb{Z}^+$?

On Wikipedia, they said that using the Weierstrass Factor Theorem, $\frac {\sin x}x$ is equal to$$\frac {\sin x}x=\left(1-\frac x{\pi}\right)\left(1+\frac x\pi\right)\left(1-\frac x{2\pi}\right)\left(1+\frac x{2\pi}\right)\left(1-\frac x{3\pi}\right)\cdots\tag1$$ Which confuses me, since I thought that the roots of $\frac{\sin x}x$ are $k\pi$ for $k\in\mathbb{Z}$ and $k\neq0$.

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    If $f(x)$ is a function, the solutions of $f(x)=0$ are not in terms of $x$, they are in terms of $x=\cdots$. The roots here, then, are not $\frac{x}{k\pi}$, then, but the values $x$ where some $1+\frac{x}{k\pi}=0$. – 2017-01-28

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$(1-\frac{x}{k\pi})= 0 $ when $x = k\pi.$