How to prove that $2^{n+3} = O(2^n)$? I am a little confused on how to prove/disprove Big O.

For the problem, $2^{n+3}= O(2^n)$, I did the following:

$$2^{n+3} \leq K \times 2^n$$

Set $K = 1$

$$2^{n+3} \leq 2^n$$

Test for large values of n (so I plugged in n = 100)

$2^{103} \leq 2^{100}$  --- which is false therefore Big O is disproven

Is this process correct?
Just, replace the constant $K = 2^3$, and you have to use this constant for your proof.

$$ 2^{n+3}  = K\times 2^n$$

So, $O(2^{n+3})  = O(2^n)$.