What is the GCD of all these numbers?

For prime $p$ and natural number $a$ ,we have $a^{p - 1} - 1$ as an integral multiple of p (by Fermat's little theorem). When p=73 then all the given numbers are of the form $a^{72}-1$ which is divisible by 73.
For 73 to be the gcd we need to show any other number, $x$ doesn't divide all of them.
Four cases arise for $x$ :
1. $0<x<73$ .
Here $x$ will not divide $x^{72}-1$ .
2. $x>73$ , and x is a prime:.
Factorise $2^{72}-1$ and $3^{72}-1$ by repeated application of identity $a²-b²=(a+b)(a-b)$ and $a³±b³=(a±b)(a²±ab+b²)$ . They don't have any other number in common except 73. Even if some other prime divides any number from the set it will not be part of GCD.
3. $x>73$ , and x is a composite:.
Since we have shown it for all primes, i.e, which are less than 73 and which are grater than 73, it suffices to say that any $x>73$ , and composite will not be part of GCD.
4. Any perfect power of 73, will also be not their GCD as 73 occurs only once in prime factorisation of $2^{72}-1$ .
So it has been shown that any other number, $x$ doesn't divide all of them.
So there GCD is $\boxed{73}$

I have the generalization for all primes. I don't know how to use TeX, so mind my formal mistakes.

We'll prove that the only nonzero and nonone number for which $2^{p-1},3^{p-1},..., (p-1)^{p-1}$ all give 1 residue is the number $p$ . (p is an odd prime).Thus, the greatest common divisor is p.

p divides all of them according to the Little Fermat Theorem. Looking at the smaller divisors, it's obvious they won't divide all of them. We note that this means that no primes between 2 and p-1 divide all of them, so any number that is divisible by them is not a solution either (won't divide all of them). So we only got to check higher numbers than p.

If we prove that no higher primes divide all of these numbers then we are done by the same argument. We'll use a little bit of polynomial theory/fields. (I swear, if anyone hasn't had a read on them I recommend it the highest for IMO or any other high level competition, it's key. It solves harder problems so easy, also really helpful in easier problems.)

Assume that $r$ is a (bigger than p) prime that divides all of our numbers. Note that $1^{p-1}$ also gives 1. Thus, the $x^{p-1}-1 \equiv 0 \pmod{r}$ polynomial has p-1 roots: $1,2,...p-1$ . It is known, that if we have a prime $p$ , then any nonzero polynomials modulo p of degree n have at max n roots (This is called Lagrange's Theorem .). It means that we can't have any other roots.

But note that since p-1 is even, r-1 is also a root, because $(r-1)^{p-1} \equiv (-1)^{p-1}=1 \pmod{r}$ , but we can't any new roots, hence $r-1$ equals a number between $1,2,..p-1$ . But that's impossible, since $r>p$ , we have $r-1>p-1$ , also $r-1>p-1>p-2>...>2>1$ so r-1 is bigger than all of them, can't equal any of them. Contradiction, no higher prime can divide all of $2^{p-1}-1,... (p-1)^{p-1}-1$ , thus only p divides it.

Finally, we have to check whether any of $p^2, p^3,...$ divides all of them. But that's trivial, we can show that $p^2$ doesn't divide $(p-1)^{p-1}-1$ because $(p-1)^{p-1}-1 \equiv -p \pmod{p^2}$ , you can see this expanding out via the binomial theorem. Hence, none of them divides it, only $p$ .

Hence their greatest common divisor is $p$ .

Ans: 73. If the power is prime, the gcd is 1, Else the gcd is the power + 1. Do from the simplest: If the power = 3, 2^3 - 1, 3^3 - 1. 7, 8 -> gcd = 1 If the power = 4, 2^4 - 1, 3^4 - 1, 4^4 - 1. 15, 80, 255 -> gcd = 5 Etc. :)

If p is a prime no. ,then (a^(p − 1) − 1) is an integral multiple of p (by Fermat's little theorem),(where a can be any integer). when p=73 then all the given no. are of the form (a^(72)-1) which is divisible by 73 (by theorem). hence gcd=73