Don't you dare try bashing this out!

Let's write the given number as $a=6^b$ , with $b>5$ . Now $6^{b-4}\equiv1\pmod5$ . Multiplying through with $6^4=1296$ we find $a=6^b\equiv 6^4 \pmod{6480}$ . Now $6480=6!\times 9$ so $a\equiv 6^4\equiv \boxed{576}\pmod{6!}$

Using Euler's theorem, we need to determine: $\displaystyle 5^{ 4^{9}} \: \text{modulo } \: \phi (720), \; 4^{9} \: \text{modulo} \: \phi(\phi (720))$

Since $\phi (720) = 192, \; \phi(192) = 64$ , we have $4^9 \equiv 0 \pmod{64}$ $5^{(\cdots)} \equiv 1 \pmod{192}$

We can't repeat these steps now because $\gcd(6,720) \neq 1$ . But $6^{(\cdots)} \equiv 6^{5^1} \equiv \boxed{576.} \pmod{720}$

This might be a solution by brute force, but still is my solution...

The main idea is we use the fact that

1. $24^4 \equiv 576 \pmod {6!}$
2. $576 = 24^2$ .

We solve it by parts using the powers.

This is $6^5$ :

$6^5 \equiv 24 \pmod {6!}$

This is ${{6}^5}^4$ :

${24}^4 = 331776 \equiv 576 \pmod {6!}$

Then we have

This is ${{{6}^5}^4}^3$ :

$576^3 = {{24}^2}^3 = {24}^6 = {24}^4 \times {24}^2$

Note that from (1) , ${24}^4 \equiv 576 \pmod {6!}$

Now, ${24}^4 \times {24}^2 \equiv 576 \times {24}^2 \pmod {6!}$

$576 \times {24}^2 = {24}^2 \times {24}^2 = {24}^4 \equiv 24^2 \pmod {6!}$

Finally, this is ${{{{6}^5}^4}^3}^2$ :

${24^2}^2 = {24}^4 \equiv 576 \pmod {6!}$

We want to find $6^{5^{4^{3^{2}}}}\ (\text{mod}\ 720)$ .

As $720=144\cdot5$ and $144$ and $5$ are coprime integers, we can rewrite the problem as finding the unique $0≤x<720$ that satisfies $6^{5^{4^{3^{2}}}}\equiv x\ (\text{mod}\ 144)$ $6^{5^{4^{3^{2}}}}\equiv x\ (\text{mod}\ 5)$

Note that $144=2^4\cdot3^2$ , thus $6^{5^{4^{3^{2}}}}\equiv0\ (\text{mod}\ 144)$ . Furthermore, $6\equiv1\ (\text{mod}\ 5)$ , and thus $6^{5^{4^{3^{2}}}}\equiv1\ (\text{mod}\ 5)$ .

So we want to find $0≤x<720$ such that $x\equiv0\ (\text{mod}\ 144)$ $x\equiv1\ (\text{mod}\ 5)$

By the Chinese Remainder Theorem, there exists one such unique $x$ . With trial and error ( $x=144,288,432,576,\ldots$ ), we find that $x=576$ .

The key is to break 6! Into coprime numbers......i found that breaking it into (16,5,9) is particularly easy. The number already contains powers of 2 and 3 which are much much higher than 4 and 2 .......so the remainder when it is divided by 16 and 9 is 0. Now when the number is divided by 5 ....it is easy to see that the remainder will be 1. So we need to find the solution of the system of congruences x= 0 mod 144 and x=1 mod 5 (forgive me for using" = "instead of congruence). Use chinese remainder theorem of division algorith to quickly find that the remainder is nothing but 144 × 4 = 576