Can someone please look at my work below? I want to make sure everything I'm doing to my inequality in my inductive step is allowable/if I need to justify along the way (for instance when I subtract $-1$ from the LHS in second to last step). Also, any advice/other tips on how I constructed this proof welcome.
$S_n:$ For every $n\in\mathbb{N}$, it follows that $2^n+1\le3^n$.
$Proof.$ We will prove this using mathematical induction.
Base Case. Let $n=1$. Then $2^n+1=2^{(1)}+1=3$ and $3^n=3^{(1)}=3$. Thus our inequality holds for $n=1$ since $3\le3$.
Inductive Step. Let the natural number $n=k\ge1$. We need to show that $S_k\Rightarrow S_{k+1}$. We use direct proof. Suppose $2^k+1\le 3^k$. Observe that
$$\begin{align*} 2^k+1 &\le 3^k \\ 2(2^k+1) &\le 3(3^k)\enspace (since\enspace (a\le b)\land( c < d) \Rightarrow ac\le bd)\\ 2^{k+1}+2&\le3^{k+1}\\ 2^{k+1}+2-1&\le3^{k+1}\\ 2^{k+1}+1&\le3^{k+1} \end{align*}$$
Thus $S_k\Rightarrow S_{k+1}$.
It follows by mathematical induction that $S_n$ is true.
