Prove the inequality $n! \geq 2^n$ by induction

In the induction step you want to show that if $k!\ge 2^k$ for some $k\ge 4$, then $(k+1)!\ge 2^{k+1}$. Since you already know that $4!\ge 2^4$, the principle of mathematical induction will then allow you to conclude that $n!\ge 2^n$ for all $n\ge 4$. You have all of the necessary pieces; you just need to put them together properly. Specifically, you can argue as follows.

Suppose that $k!\ge 2^k$, where $k\ge 4$; this is your induction hypothesis. Then $$\begin{align*} (k+1)! &= (k+1)k!\text{ (by the definition of factorial)}\\ &\ge (k+1)2^k\text{ (by the induction hypothesis)}\\ &> 2\cdot2^k\text{ (since }k\ge 4\text{)}\\ &= 2^{k+1}. \end{align*}$$ This completes the induction step: it shows that if $k\ge 4$, then $$k!\ge 2^k \implies (k+1)!\ge 2^{k+1}.$$

I am going to provide a different way of going about it--this is essentially Brian M. Scott's proof in reverse. As people have pointed out, if you can show that $n! > 2^n$, then you will have shown that $n! \geq 2^n$ (in this sense, you can think of $>$ as stronger than $\geq$ because $>$ implies $\geq$).

When considering $n! \geq 2^n$ and how you got stuck trying to work from left to right to prove the argument by induction, it may behoove you in some instances to actually work from right to left since $n! \geq 2^n$ is the exact same as $2^n \leq n!$. With this in mind, I will give a small proof that $2^n < n!$ for $n\geq 4$ (like I said, it is very similar to Brian's, but it provides a different way of going about it nonetheless that may be useful for you or others in the future).

For $n\geq 4$, denote the statement involving $n$ by $$ S(n) : 2^n<n!. $$ Base step ($n=4$): Since $2^4=16$ and $4!=24$, the statement $S(4)$ is true.

Inductive step: Fix some $k \geq 4$ and assume that $$ S(k) : 2^k < k! $$ is true. To be shown is that $$ S(k+1) : 2^{k+1} < (k+1)! $$ follows. Beginning with the left side of $S(k+1)$, \begin{align} 2^{k+1} &= 2(2^k)\\[0.5em] &< 2(k!)\tag{by $S(k)$}\\[0.5em] &< (k+1)(k!)\tag{since $k\geq 4$}\\[0.5em] &= (k+1)!, \end{align} the right side of $S(k+1)$. This concludes the inductive step $S(k)\to S(k+1)$.

Thus, by mathematical induction, for all $n\geq 4$, the inequality $S(n)$ is true.