Solving a Binomial Expansion Question I have a question which asks to find the coefficient of x and the constant term, for $f_n(x)$ given that
$f_1(x) = (x - 2) ^ 2$ and
$f_{n+1}(x) = (f_n(x) - 2) ^ 2, n >= 1$
Now I tried to derive the values like solving for $f_2$ and $f_3$ , but its getting too long. What is a shorter way to solve these problems?
I guess that the constant term would be $4$ or $4^{(n-1)!}$ but I am not sure.
I was proceeding like this,
$f_2 = (f_1 - 2) ^ 2 $
so for $-4 * f_1$ the constant term would be -16
and for $f_1^2$ it will be 16 for $2^4$ and $2*f_1$ would be 4 . so in total that's +4 .. so it could be 4 or $4^{0!}$.  

At $f_2$, const term is $(2^2-2)^2=2^2=4$, So, for $f_3$, you get the same thing, and this goes on for every $n$ . 
Now the constant term being fixed, notice $x$ ca be formed only by multiplying a constant term with a $x$ containing term.So we need only to consider the $2\times (term\ with \ x) \times 2$ while squaring.n So, notice the sequence..
$$f_1\ ->(-2)$$
$$f_2\ ->2\times (-2)\times 2=(-2)\times 2^2$$
$$f_3\ ->2\times (-2)2^2\times 2=(-2)\times 2^4$$
$$f_4\ ->2\times (-2)2^4\times 2=(-2)\times 2^6$$
$$\vdots$$
$$f_n\ ->2\times (-2)2^{2n-4}\times 2=(-2)\times 2^{2n-2}$$
