Prove $2^{2n+7} \equiv 2 (mod 3)$ for all integer and non-negative number n. Originally, I want to prove this problem:
for all integer and non-negative number $n$,
$$\frac{1}{3}\times2^{2n+7}+\frac{37}{3} \equiv 0\quad(\text{mod}\space3)$$
I tried like this:
$$\frac{1}{3}\times2^{2n+7}+\frac{37}{3} = \frac{1}{3}(2^{2n+7}+37)$$
therefore $2^{2n+7}+37$ must be $\equiv0\space(\text{mod}\space3)$ and greater than or equal to $3$.
Since $2^{2n+7}+37$ is bigger than $37$(for all $n$), we can ignore the latter.
By the way, $37 \equiv 1(\text{mod}\space3)$.  Therefore $2^{2n+7} \equiv 2(\text{mod}\space3)$.
I stuck here. is there any way to prove this?
and following calculation correct?
$$2^{2n+7} =  2^7\times 2^{2n}\quad \text{and}\quad 2^7 \equiv 2 (\text{mod}\space3)$$Therefore,
$$2^{2n+7}  =2^7\times 2^{2n} \equiv  2\times 2^{2n} (\text{mod}\space3)$$

Thanks Will199 and Evariste.
for all integer and non-negative number $n$,
$2^{n}\equiv (-1)^{n} \equiv -1($when n is odd$)$ or $1$(when n is even) $(mod \quad3)$
$2n+7$ is always odd because $2n$ is even and $7$ is odd and even+odd=odd.
So, $2^{2n+7}\equiv (-1)^{2n+7} \equiv -1 \equiv 2 \quad(mod \quad3)$.
