Proving $x + \sin x + \sin(x + \sin x) + \sin(x + \sin x + \sin(x + \sin x)) + ... = \pi $

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I was given a very interesting sum of series by a friend. The problem is to prove that if $ x \in (0, 2\pi) $ , then

$$ x + \sin x + \sin(x + \sin x) + \sin(x + \sin x + \sin(x + \sin x)) + ... = \pi $$

The $n^{th}$ term is related to the sum of all the previous $n-1$ terms as

$$ T_n = \sin ( S_{n-1} ) $$

I tried doing something using this relation, which didn't yield anything significant.
I also tried using Euler's form of a complex number to somehow convert it into a more manageable series, which also failed miserably.
Any hints or approaches would be appreciated.
The problem is supposed to be at the high school level.