Elementary proof that the MacLaurin series of $\sin x$ converges to $\sin x$ for all $x$ In my book it is given: 
$\sin x = x- \dfrac {x^3}{3!}+\dfrac{x^5}{5!}- \dfrac{x^7}{7!}...$
I googled around for a proof but couldn't understand any of them. I would like to know if there's any elementary high school level proof the series

I faced same problem here. We know that series approximated around $ x= 0$ and it converged but how do we know that it converge to $\sin x $ ?
As limited knowledge in math. I prefered to start with Euler's formular see more in  wikipedia
$$e^{i\theta} =\cos \theta + i\sin \theta $$
Which can be proof as shown in wiki's page by
$$ f(\theta) = e^{-i\theta}(\cos \theta + i \sin \theta)$$
$$ f'(\theta) = e^{-i\theta}(-\sin \theta + i \cos \theta) + -ie^{-i\theta}(\cos \theta + i \sin \theta) = 0$$
And
$$ f(0) = e^{-0i}(\cos 0 + i \sin 0) = 1$$
As $f'(\theta) = 0$ then $f(\theta)$ is constant. Then
$$ f(\theta) = 1$$
$$ e^{-i\theta}(\cos \theta + i \sin \theta) = 1$$
$$ \cos \theta + i \sin \theta = e^{i\theta}$$
Now, we look at power factor of $e^x$
$$ e^x = \sum_{n=0}^{\infty} \frac {x^n} {n!}$$
Then
$$ e^{ix} = 1 + ix + \frac {(ix)^2} {2!} + \frac {(ix)^3} {3!} + \frac {(ix)^4} {4!} + \frac {(ix)^5} {5!}+...+ \frac {(ix)^n} {n!}$$
$$ e^{ix} = 1 + ix - \frac {(x)^2} {2!} - \frac {(ix)^3} {3!} + \frac {(x)^4} {4!} + \frac {(ix)^5} {5!}-...\pm \frac {(ix)^n} {n!}$$
$$ e^{ix} = (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!}-...) +  i(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}-...) $$
Just for sin part, you see that
$$sinx = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}-...$$
Now the only thing you need to proof is power factor of $e$ which easier to find the proof and again, with my limited knowledge of math I will go wih this.
$$ \frac d {dx} e^x = e^x$$
And.
$$ \frac d {dx} e^x = \sum_{n=0}^{\infty} \frac d {dx} \frac {x^n} {n!}$$
$$ \frac d {dx} e^x =  0 + \sum_{n=0}^{\infty} \frac d {dx} \frac {x^n} {n!}$$
$$ \frac d {dx} e^x = \sum_{n=0}^{\infty} \frac {nx^{n-1}} {n!}$$
$$ \frac d {dx} e^x = \sum_{n=0}^{\infty}  \frac {x^{n-1}} {(n-1)!} = \sum_{n=0}^{\infty} \frac {x^n} {n!}$$
As derivative of series equal to itself for all $x$ and For $x = 0,\sum_{n=0}^{\infty} \frac {x^n} {n!} = 1$
which means power series of $e^x$ is valid for all of x
