Is there a formula for $n^{th}$ power of $\cos(ax+b)$? Is there a closed form formula or trigonometric relation for $\cos^{n}(ax+b)$?
 A: Let's forget $a$ and $b$, as it's not the problem here.
$$\cos^n x=\left(\frac{e^{ix}+e^{-ix}}{2}\right)^n=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}e^{kix}e^{-(n-k)ix}\\=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}e^{(2k-n)ix}\\=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}(\cos(2k-n)x+i\sin(2k-n)x)\\
=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}\cos(2k-n)x$$
A: As mentioned in your comment, if you are only looking to find the period of $cos^n(ax+b)$, it wouldn't be necessary to write a closed form formula or expression for it.
When $n$ is even, We see that $(\cos(\pi - (ax+b)))^n = (\cos(\pi + (ax+b)))^n$ and this turns the period of the function into $\pi/a$, whereas when $n$ is odd, we don't have this luxury anymore. Instead we have $\cos^n(ax+b)$ going its full $2\pi/a$ cycle.
Edit: As @Rob Arthan pointed out in his comment, given $f(x)$ with period $T$ and some function $g(x)$; $h(x) = g(f(x))$ has period $T/k$ for some integer $k$. It is also possible to observe that $T$ is a period of $h$, but often depending upon the function $g$ we can do better.
In our case $f = \cos(ax+b)$ and $g(x) = x^n$. When $n$ is odd, $g(x)$ is injective, and so every input to $g$ has a unique associated output, which means there is no room in $[0,2\pi/a]$ (the original period of $f$) for repetition of the same output values. This is why, when $n$ is odd, the period of $h$ is the same as the period of $f$. However, when $g$ is not injective, we have multiple inputs being mapped to the same output. In the case of an even $n$, we have exactly two inputs in a given period being mapped to the same output, which essentially means we are repeating the graph of $f$ twice in $[0,2\pi/a]$ upon composing, which means the period is halved simply giving us a period of $\pi/a$.
