Is $72!/36! -1$ divisible by 73? Is $\frac{72!}{36!}-1$ divisible by the number 73?

I am not getting a clue in which direction should I go, though I did small amount of work by converting the above expression in the below given form

$$(1.3.5.7.9...69.71).2^{36} - 1$$

and

$$36!\binom{72}{36} -1$$

I am unable to proceed further.
Let $p=73$. $p$ is a prime number. Thus by Wilson's theorem 
$$72! = (p-1)!\equiv -1 \mod p$$

Notice that $\forall 1\le n\le 36$, $73-n \equiv -n \mod p$, thus

$$ \frac{72!}{36!}\equiv (37\cdot 38\cdots 72)\equiv (-1\cdot -2\cdots -36) \equiv (1\cdot 2\cdots 36)(-1)^{36} \equiv (1\cdot 2\cdots 36) \equiv 36! \mod p$$

Suppose for the sake of contradiction that $ \frac{72!}{36!} \equiv 1 \mod p$, then $ 36! \equiv 1 \mod p$ and 

$$ 72! \equiv (36!)^2 \equiv 1 \mod p$$

which is not true.

So we can conclude that $\frac{72!}{36!} \mod p$ is not $1$.
Not a very glorious method, but it can be verified by hand. The numbers on the right are the prefix products of those on the left, modulo $73$ (e.g. $37\cdot38\bmod73=19$).

$$\begin{align}37&amp;\to37\\
38&amp;\to19\\
39&amp;\to11\\
40&amp;\to2\\
41&amp;\to9\\
42&amp;\to13\\
43&amp;\to48\\
44&amp;\to68\\
45&amp;\to67\\
46&amp;\to16\\
47&amp;\to22\\
48&amp;\to34\\
49&amp;\to60\\
50&amp;\to7\\
51&amp;\to65\\
52&amp;\to22\\
53&amp;\to71\\
54&amp;\to38\\
55&amp;\to46\\
56&amp;\to21\\
57&amp;\to29\\
58&amp;\to3\\
59&amp;\to31\\
60&amp;\to35\\
61&amp;\to18\\
62&amp;\to21\\
63&amp;\to9\\
64&amp;\to65\\
65&amp;\to64\\
66&amp;\to63\\
67&amp;\to60\\
68&amp;\to65\\
69&amp;\to32\\
70&amp;\to50\\
71&amp;\to46\\
72&amp;\to27\\
\end{align}$$