Induction proofs in Big-O notation
We need to inspect the factors after that, since the factors of $n!$ grows linearly while the factors of $2^n$ stays constant. We use the first four factors and the rest of them to formulate the constant factors used for the big-O proof. It is obvious that $2^4 = 16$ while $4! = 24$. However, by observing the factors, we notice $2^n$ has $16$ as a factor and $n!$ has $24$ as a factor for all $n \geq 4$ and we factor out the first four [the first $N$] to clearly highlight that the factorial grows faster.

The factors of $\frac{1}{16} 2^n$ factors out the first $2\cdot 2 \cdot 2 \cdot 2$. It is equal to $2\cdot 2\cdot 2\,\cdot \, ... \cdot \, 2$ with $(n-4)$ factors. The factors of $\frac{1}{24}n!$ factors out $1 \cdot 2 \cdot 3 \cdot 4$ and is equal to $5 \cdot 6 \cdot 7\, \cdot \,... \cdot \, n$, again with $(n-4)$ factors. When $n = 4$ then these factored-out functions are equal to unity. In fact:


$2\cdot 2\cdot 2\,\cdot \, ... \cdot \, 2\cdot 2 = c_f2^n$ where $c_f = \frac{1}{16}$ whereas
$5 \cdot 6 \cdot 7\, \cdot \,... \cdot \, (n-1) \cdot n = c_gn!$ where $c_g = \frac{1}{24}$.


The $(n-4)$ factors of $c_f2^n$ is clearly less than the $(n-4)$ factors of $c_gn!$. We arrive at $c_f2^n &lt; c_gn!$. Move the constant factor on the LHS to obtain $2^n &lt; \frac{c_g}{c_f}n! = \frac{24^{-1}}{16^{-1}}n! = \frac{16}{24}n! = \frac{2}{3}n!$.

By inspecting the factors of $2^n$ and $n!$ after the initial four, and knowing that for all $n &gt; 4:\: 2^n &lt; n!$, we factored out the first four terms of each function and inspected the rest to determine $2^n \leq \frac{2}{3}n!$.

The formal definition of big-O states that:

$$\exists c &gt; 0\; \exists N &gt; 0: \forall n\; (n&gt;N) \Longrightarrow f(n) \leq cg(n)$$

In your case, $c = \frac{2}{3} = \frac{16}{24}$ and $N = 4$.