Proving $n! \ge 2^{n-1 }$for all $n\ge1 $by mathematical Induction [duplicate]

Im trying to solve the following question

In the second step where do they get $k!=2^k-1?$

In the induction step, they say "Suppose that $k! \ge 2^{k-1}$ for some $k \ge 1$."

That then allows them to say "$(k+1)\,k! \ge (k+1)\,2^{k-1}$", multiplying both sides of the inequality by the positive $(k+1)$.

Consider this example...similar to your problem..

Prove $n!>2^n$ $(n \geq4)$ Prove it at first for any $n \geq 4$.

Assume it true for some $k$. let it be true $k!>2^k$

Now we try to prove it true for $k+1$. We have to prove $(k+1)!>2^{k+1}$.

We know $2^{k+1}$=$2^k*2$.

Also,we know $(k+1)!=k!*(k+1).$ We also assumed earlier $k!>2^k$.

So, we can write $k!(k+1)>2^k*2$ [Since,$k>4 ,k+1>4>2$ which is in the RHS and $k!>2^k$ as we assumed earlier].

Hope this helps you to understand the problem and solve it in a similar way.