Question about Chebyshev Polynomials in Beardon

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I happen to be reading through Beardon's book, Iteration of Rational Functions, and I have come across a statement I don't quite believe. He uses it a little later on, so I'm concerned with clearing it up before I move on. First he makes the definition

$$ T_k(\cos (x)) = \cos (kx). $$

This is fine, this is the usual definition. But then he states that from this, we can see

$$ (T_k)^n(x) = \cos (k^n x). $$

This is what I'm having trouble seeing. For example, I know

\begin{align} T_2(\cos (x)) &= \cos (2x) = 2 \cos^2(x) -1\\ T_4(\cos (x)) &= \cos (4x) = 8\cos^4(x)-8\cos^2(x)+1. \end{align}

Using these identities, then according to his formula

$$ (T_2)^2(\cos (x)) = \cos (4x). $$

But it is easily seen that

$$ (T_2)^2(\cos (x)) = (2 \cos^2(x)-1))^2 = 4\cos^4(x)-4\cos^2(x)+1, $$ while

$$ \cos(4x) = 8\cos^4(x)-8\cos^2(x)+1. $$

Can someone clarify for me? Thank you in advance.