Mathematics Induction $3^n-2\ge2^n-1$ Can someone help me to answer this question?

$3^n-2\ge2^n-1$

Prove that $n$ works for all positive integers with mathematic induction.
I tried to answer it but I think I made a mistake.
This is my answer
$3^n-2\ge2^n-1$
$P(1)$
Will be proven right
$3^1-2\ge2^1-1$
$3-2\ge2-1$
$1\ge1$
Proven Right
$P(k)$
Will be assumed right
$3^k-2\ge2^k-1$
$P(k+1)$
Will be proven right
$3^{k+1}-2\ge2^{k+1}-1$
Prove :
using $P(k)$
$3^k-2\ge2^k+-1$
after this I'm blanking out
 A: As mentioned in the comments, you made expansion mistakes.
An induction proof usually has the basic structure of showing the base case, assuming the "inductive hypothesis" that the previous cases are true, and then showing that the next case is true. I struggled to read that in your working, so here is a proof:
Base case: $3^1-2=1\ge2^1-1$ - we can induct from $n=1$ upwards.
Assume $3^n-2\ge2^n-1\implies3^n\ge2^n+1$ (inductive hypothesis). Then$3^{n+1}=3\cdot3^n\ge3(2^n+1)=3\cdot2^n+3$ by the inductive hypothesis.
$$3^{n+1}\ge3\cdot2^n+3\gt2\cdot2^n+3=2^{n+1}+3\\3^{n+1}\gt2^{n+1}+3\implies3^{n+1}\gt2^{n+1}+1\\\therefore3^n-2\ge2^n-1\implies3^{n+1}-2\gt2^{n+1}-1$$
And so for any $n$ for which the previous case is true, the next case of $n+1$ is also true. Apply this to $n=1$, and you are done.
A: You may use binomial theorem:
$$3^n-2=(2+1)^n-1=2^n-1+ P(2, n)$$
where $p(2, n)$ is a positive polynomial in terms of 2 and n, therefore:
$$ 3^n-2\ge 2^n-1$$
.If this is true for n it must also be true for (n+1):
$$3^{n+1}-2=(2+1)^{n+1}-1=2^{n+1}-1+ P(2, n+1)$$
$$\Rightarrow 3^{n+1}-2\ge 2^{n+1}-1$$
which is obvious,
