Growth of exponential functions according to the big O notation
You swapped two functions. Notice that:
$$n^{\sqrt{n}} = 2^{\sqrt{n} \log n } = o(2^{\frac{n}{2}})$$.

Once this is fixed, the factorial is in the correct order since very rough inequalities show that 
$$n! = n \cdot (n-1) \cdot \ldots \cdot 2 \ge \underbrace{2 \cdot 2 \cdot \ldots \cdot 2}_{n-1 \text{ times}} = 2^{n-1} = \omega(2^{\frac{n}{2}}),$$
and
$$n =  n \cdot (n-1) \cdot \ldots \cdot 2 \le \underbrace{n \cdot n \cdot \ldots \cdot n}_{n-1 \text{ times}} = n^{n-1} = o(n^n).$$

In general, from Stirling's approximation you know that $n! = \Theta( \sqrt{n} \cdot (n/e)^n )$.