How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$? 
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.

Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots$$
So, $$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$
However I'm stucked from here on. Thank you in advance!

If you want to use the Maclaurin series for $\cos$ in order to obtain the first few terms of the Maclaurin series for $\cos^6$ proceed as follows: Start with
$$\cos x=1-{x^2\over2}+{x^4\over24}+?x^6=1-{x^2\over2}\left(1-{x^2\over12}+?x^4\right)\ .$$Using the binomial theorem you then obtain
$$\cos^6 x=1-{6\choose 1}{x^2\over2}\left(1-{x^2\over12}+?x^4\right)+{6\choose2}{x^4\over4}(1+?x^2)+?x^6\ .$$
Now collect terms:
$$\cos^6 x=1-3x^2+4x^4+?x^6\ .$$
