How to show $\lceil ( \sqrt3 +1)^{2n}\rceil$ where $n \in \mathbb{ N}$ is divisible by $2^{n+1}$ 
Show that $ \left\lceil( \sqrt3 +1)^{2n}\right\rceil$ where $n \in  \mathbb{N}$ is divisible by $2^{n+1}$.


I wrote the binomial expansion of $ ( \sqrt3 +1)^{2n}$ and  $( \sqrt3 -1)^{2n}$ and then added them to confirm that the next integer is even. Afterwards I applied $AM \ge GM$ on the two terms to get $ ( \sqrt3 +1)^{2n} + ( \sqrt3 -1)^{2n} \ge (2^{n+1})$.
Now I'm unable to figure out the next step. Any help would be appreciated. :)

Let:
$$ A_n = (\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n} = (4+2\sqrt{3})^n+(4-2\sqrt{3})^{n}.\tag{1} $$
Since $0<4-2\sqrt{3}<\frac{2}{3}$, we have that $A_n$ is the integer giving the ceiling of $(\sqrt{3}+1)^{2n}$.
Now we have: 
$$ A_0 = 2, \quad A_1 = 8,\qquad A_{n+2} = 8 A_{n+1} - 4 A_{n}\tag{2} $$
and if we take:
$$ \nu_2(n) = \max\{m\in\mathbb{N}: 2^m\mid n\} \tag{3}$$
it happens that:
$$ \nu_2(A_n)\geq n+1\tag{4} $$
can be proved by induction from $(2)$. We can also factor a $2^n$ from the RHS of $(1)$ then just study the parity of the sequence given by:
$$ B_n = (2+\sqrt{3})^n+(2-\sqrt{3})^n,\tag{5}$$
for which:
$$ B_0=2,\quad B_1=4,\quad B_2=14,\quad B_{n+2}=4B_{n+1}-B_n\equiv -B_n\pmod{4}.\tag{6}$$
That gives:

$$ \nu_2(A_n) = \left\{\begin{array}{rl}n+1 & \text{if }n\text{ is odd,}\\n+2 &\text{if }n\text{ is even.}\end{array}\right.\tag{7}$$

