Inductive proof - Inequality

Question

I am trying to do mathematical induction to prove an inequality. However, I got stuck in the middle of the proof. I have started my proof from the inductive step $3^{k+1}=3^k*3 ≥3k*3$. I am stuck in here. The below picture is about what I did. I also tried in a separate way. Picture number $2$ is the alternate way that I tried to prove my equation. The main question is attached to image $1$. Using mathematical induction, prove:
$〖3n≤3〗^n$ for all natural numbers $n$

Can anyone please help me out with that?
Here is the snip of the problem(First attempt)
Second attempt

Answer 1

\begin{gather*} 3^{n} >3n\\ 3^{n} +3 >3( n+1)\\ Now,\ 3^{n+1} >3^{n} +3,\ \\ because\ 2\times 3^{n} >3\ for\ all\ n.\\ Hence,\ 3^{n+1} >3( n+1) ,\ hence\ proved.\\ \\ \end{gather*}

Answer 2

Since you are told to prove by induction, I will proceed by induction on $n$

For the base case we know for $n=1$, $3^1 = 3$. Thus assume the assertion holds for $n=k$ with $1 \leq k \leq n$. Consider $n=k+1$,

We have $3^{k+1} = 3^k.3 \geq 3k.3 = 3k+3k+3k > 3k+3 = 3(k+1)$ establishing $n = k+1$.