Only Stirling?

We need the following approximation formulas for $n!$ ,

1. Burnside's formula: $\displaystyle\ n!\approx \sqrt{2\pi}\left ( \frac{n+\frac{1}{2}}{e} \right )^{n+\frac{1}{2}}$
2. Gosper's formula: $\displaystyle\ n!\approx\ n^{n}e^{n}\sqrt{\pi}\sqrt{2n+\frac{1}{3}}$
3. Necdet Batir's formula: $\displaystyle n!\approx\ n^{n}e^{-n}\sqrt{2\pi \left (n+\frac{1}{6}+\frac{1}{72\left ( n+\frac{31}{90} \right )}-\frac{5929}{2332800\left ( n+\frac{3055123}{11205810} \right )^{3}} \right )}$
4. And of course, Stirling's formula.

We can rewrite the limit as: $\displaystyle \lim_{n}\frac{\left (n! \right )^{4}}{\sqrt{2\pi}\left (\frac{n+\frac{1}{2}}{e} \right )^{n+\frac{1}{2}}n^{n}e^{-n}\sqrt{\pi}\left (\sqrt{2n+\frac{1}{3}} \right )n^{n}e^{-n}\sqrt{2\pi \left (n+\frac{1}{6}+\frac{1}{72\left ( n+\frac{31}{90} \right )}-\frac{5929}{2332800\left ( n+\frac{3055123}{11205810} \right )^{3}} \right )}\left (n^{n}e^{-n}\sqrt{2\pi n} \right )}$ It's easy to see the aproximation formulas for each $n!$ , hence the sequence converges to $\boxed{1}$ .

Note: The 4th formula is VERY powerful & accurate. For example, let's take $n=6.$ Okey, $6!$ is clearly $720$ and the Necdet Batir's formula give us $719.999999376$