Consider the following calculation:
$$16!=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15\cdot16$$ $$\phantom{16!}\leq1\cdot2\cdot4\cdot4\cdot8\cdot8\cdot8\cdot8\!\cdot\!16\!\cdot16\cdot16\cdot16\cdot16\cdot16\cdot16\cdot16$$ $$=2^1\cdot4^2\cdot8^4\cdot16^8$$ $$=(2^1)^{2^0}\cdot(2^2)^{2^1}\cdot(2^3)^{2^2}\cdot(2^4)^{2^3}$$ $$=2^{(1\cdot2^0+2\cdot2^1+3\cdot2^2+4\cdot2^3)}$$ $$=2^{49}$$
In general, the factorial of a power of two is bounded by
$$(2^n)!\leq2^{(1\cdot2^0+2\cdot2^1+\cdots+n\cdot2^{n-1})}.$$
Here we'll use the identity $1+2x+3x^2+\cdots+nx^{n-1}=(1-(n+1)x^n+nx^{n+1})/(1-x)^2$, which can be gotten by differentiating $1+x+x^2+\cdots+x^n=(1-x^{n+1})/(1-x)$, thus:
$$(2^n)!\leq2^{(1-(n+1)2^n+n2^{n+1})}$$ $$=2^{(1+(n-1)2^n)}$$ $$=2\cdot\left(\frac{2^n}{2}\right)^{2^n}$$
Now consider the "opposite" calculation:
$$15!=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15$$ $$\phantom{16!}\geq1\cdot2\cdot2\cdot4\cdot4\cdot4\cdot4\cdot8\cdot8\cdot\,8\;\cdot\;8\;\cdot\;8\;\cdot\;8\;\cdot\;8\;\cdot\;8$$ $$=1^1\cdot2^2\cdot4^4\cdot8^8$$ $$=(2^0)^{2^0}\cdot(2^1)^{2^1}\cdot(2^2)^{2^2}\cdot(2^3)^{2^3}$$ $$=2^{(0\cdot2^0+1\cdot2^1+2\cdot2^2+3\cdot2^3)}$$ $$=2^{34}$$
In general we have
$$(2^n-1)!\geq2^{(1\cdot2^1+2\cdot2^2+\cdots+(n-1)\cdot2^{n-1})},$$
which simplifies thus:
$$=2^{2(1\cdot2^0+2\cdot2^1+\cdots+(n-1)\cdot2^{n-2})}$$ $$=2^{2(1-n2^{n-1}+(n-1)2^n)}$$ $$=2^{(2+(n-2)2^n)}$$ $$=4\cdot\left(\frac{2^n}{4}\right)^{2^n}$$
These two results suggest the inequality
$$4\cdot\left(\frac{n}{4}\right)^n\quad\leq\quad(n-1)!\quad\leq\quad n!\quad\leq\quad2\cdot\left(\frac{n}{2}\right)^n$$
We've shown that it's true when $n$ is a power of two. Is it true for all $n$?
For the upper bound, we can use the AM-GM inequality. First we consider the case where $n=2m$ is even:
$$n!=1\cdot2\cdot3\cdots(m-1)\cdot m\cdot(m+1)\cdots(2m-2)\cdot(2m-1)\cdot(2m)$$ $$=1(2m-1)\cdot2(2m-2)\cdot3(2m-3)\cdots(m-1)(m+1)\cdot m\cdot(2m)$$ $$\leq\quad(m)^2\quad\cdot\quad(m)^2\quad\cdot\quad(m)^2\quad\cdots\quad(m)^2\quad\cdot\quad m\cdot(2m)$$ $$=(m^2)^{m-1}\cdot m\cdot(2m)$$ $$=2\cdot m^{2m}$$ $$=2\cdot\left(\frac{n}{2}\right)^n$$
Next we consider the case where $n=2m+1$ is odd:
$$n!=1\cdot2\cdot3\cdots(m-1)\cdot m\cdot(m+1)\cdots(2m-1)\cdot(2m)\cdot(2m+1)$$ $$=1(2m)\cdot2(2m-1)\cdot3(2m-2)\cdots m(m+1)\cdot(2m+1)$$ $$\leq\;(m+\tfrac12)^2\;\cdot\;(m+\tfrac12)^2\;\cdot\;(m+\tfrac12)^2\;\cdots\;(m+\tfrac12)^2\;\cdot\;(2m+1)$$ $$=\big((m+\tfrac12)^2\big)^m\cdot(2m+1)$$ $$=2\cdot(m+\tfrac12)^{2m+1}$$ $$=2\cdot\left(\frac{n}{2}\right)^n$$
For the lower bound, we can use induction. Suppose the bound is valid, $(n-1)!\geq4(n/4)^n$, for some number $n$. Then for the next number we have
$$n!=n\cdot(n-1)!$$ $$\geq n\cdot4\left(\frac{n}{4}\right)^n$$ $$=4\cdot\frac{n^{n+1}}{4^n}$$ $$\overset?\geq4\cdot\left(\frac{n+1}{4}\right)^{n+1}$$ $$=\frac{(n+1)^{n+1}}{4^n}$$
so the result will be established if we can get that questioned inequality:
$$4\cdot n^{n+1}\overset?\geq(n+1)^{n+1}$$ $$\left(1-\frac{1}{n+1}\right)^{n+1}\overset?\geq\frac{1}{4}$$
This is seen to be true, e.g. here: Prove that $ \left(1-\frac{1}{n}\right)^n > \frac{1}{6} $ for $n\geq 2$