Exponent of 3

Relevant wiki: Divisibility Rules (2,3,5,7,11,13,17,19,...)

Yes , divisibility rules state that a positive integer $N$ is divisible by 9, if its sum of digits is divisible by 9 (see proof). Since all $3^n$ , where $n \ge 2$ are divisible by 9, their sums of digits are also divisible by 9.

Proof: Any $n$ -digit positive integer can be expressed as:

$\begin{aligned} N & = \sum_{k=0}^{n-1} 10^k d_k = \sum_{k=0}^{n-1} (9+1)^k d_k & \small \color{#3D99F6} \text{where }d_k \text{ is the }k\text{th digit of }N. \\ \implies N & \equiv \sum_{k=0}^{n-1} (9+1)^k d_k \equiv \sum_{k=0}^{n-1} d_k \text{ (mod 9)} \end{aligned}$

Note that $\displaystyle \sum_{k=0}^{n-1} d_k$ is the sum of digits of $N$ . Therefore, $N$ is divisible by 9, if its sum of digits is divisible by 9.

It's simple. Let us take a number ${\color{#E81990}3^k} \space (k\in \mathbb N > 1)$

Now the above number can be expressed as : $\begin{aligned} {\color{#E81990}3^k} = {\color{#20A900}3^2} \times {\color{#3D99F6}3^{k - 2}} \\ \text{(or)} \quad \quad \\ {\color{#E81990}3^k} = {\color{#20A900}9} \times {\color{#3D99F6}3^{k - 2}} \space \space \\ \end{aligned}$ So, the number ${\color{#E81990}3^k} \space (k\in \mathbb N > 1)$ is divisible by $\color{#20A900}9$ . Now, a number is divisible by $\color{#20A900}9$ if and only if the sum of digits of that number is also divisible by $\color{#20A900}9$ . So, the sum of digits of number ${\color{#E81990}3^k} \space (k\in \mathbb N > 1)$ is always a multiple of $\color{#20A900}9$ .

Any 3^k (k>1), divides by 9 = 3^2 since k>=2

Quite simply if $k \geq 2$ , then we can rewrite $3^k = 9\times 3^s$ where $k = 2+s$ . This means that $3^k$ is divisible by $9$ $\forall k \geq 2$ .

Divisibility rule for $9$ :

If the sum of the digits of $n$ is divisible by $9$ then $9$ divides $n$ .

Since $9$ divides $3^k$ , $9$ must also divide the sum of the digits of $3^k$ .

If the sum of all digits is divisible by 9, that means that the number is divible by 9.

3^2 = 9

And :

    3^k = 3^2 × 3^l 

           = 9 × 3^l

with k > 1 and l an integer

Thus 3^k is always divisible by 9 for any k > 1