I am trying to prove a something regarding Chebyshev polynomials. Given the polynomials $T_n(x), n = 0, 1, \ldots$ which are recursively defined by $\begin{cases} T_0(x) = 1\\ T_1(x) = x \\T_n(x) = 2x T_{n−1}(x) − T_{n−2}(x), & \text{for } n \geq 2\end{cases}$
I want to show that
For every $n$, $T_n(x) = \cos(n \arccos(x))$
Is there some type of proof I could use or is just plugging in values the only way?