Proof by induction that $f(n) = 1-2^{2^n}$, where $f(0) = 3$ and $f(n) = 2 f(n-1) - (f(n-1))^2$ I am doing a textbook question which state that a function $f:\mathbb{N}\to\mathbb{Z}$ is a recursively defined as shown bellow
$f(0) =3$,
$f(n) = 2\cdot f(n-1) -(f(n-1))^2 $ if $n\ge1$.
Prove that $f(n) = 1-2^{2^n}$ for all integer $n\ge1$.
I am trying to prove this by induction, so I started by using my base step :
Let $n=1 : f(1) = -3 $
then I did the inductive step for $n = k-1$
which gave me $f(k-1) = 1-2^{2^{k-1}}$ but I don't know how to proceed from here.. can one explain. thanks

$$\begin{align}f(k)&=2f(k-1)-(f(k-1))^2\\&=2 \left(1-2^{2^{k-1}}\right)-\left({1-2^{2^{k-1}}}\right)^2\\&=2-2^{2^{k-1}+1}-\left(1-2^{2^{k-1}+1}+2^{2^k}\right)\\&=1-2^{2^{k}}.\end{align}$$
P.S.
$$2\left(1-2^{2^{k-1}}\right)=2-2^1\cdot 2^{2^{k-1}}=2-2^{1+2^{k-1}}=2-2^{2^{k-1}+1}.$$
$$\left(1-2^{2^{{k-1}}}\right)^2=1^2-2\cdot 1\cdot 2^{2^{k-1}}+\left(2^{2^{k-1}}\right)^2=1-2^{2^{k-1}+1}+2^{2^{k-1}\times 2}.$$
