Proof of $n!\geq2^{n-1}$ by mathematical induction

\begin{align} (k+1)! &= k!(k+1) \\ &\ge2^{k-1}(k+1) \end{align}

Can you complete it?

You're off by $1$ in your basis observations; you should have, instead, the following:

$$ 1! \geq 2^0 \\ 2! \geq 2^1 \\ 3! \geq 2^2 $$

For the induction step, you should be thinking like this:

$$ 2^k = 2 \times 2^{k-1} \leq \cdots $$

Can you complete that line?

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P.S. Siong Thye Goh's answer gives the other half... :-)

Then for $n = k+1$, we have $$2^k = 2\cdot2^{k-1} \le 2 \cdot k!\ \text{ by inductive argument}$$ From here, it is easy to see that $2\cdot k! \le (k+1)k! = (k+1)!$