9876543210!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Or what if you restricted the number of symbols to be no more than 20 symbols total.

10!^2!^3!^4!^5!^6!^7!^8!^9!

HA! 2^3!^4!^5!^6!^7!^8!^9!^10! ;-) (2! = 2) and (2^x) > (x ^2) where x > 3.

Then again 2^3^4^5^6^7^8^9^(10!!!!!!!!!) is larger. Also if it's it's simply a question of symbols 9!! is larger than 9^9 so 9 followed by n ! is probably your best bet without going into higher level math like BB(BB(...(9))...)

10→9→8→7→6→5→4→3→2

Somehow I think that Conway's chained arrow notation is cheating.

9876543210

If I'm also allowed an operator, 3↑⁹⁸⁷⁶⁵⁴¹⁰2

ha, yes allowed operators, as many as youd like.

Then the best I can come up with on no notice is 2 ↑^(4 ↑^(6 ↑^(8↑^(9↑10)) 7) 5) 3.

That would be trivial to increase substantially by using some other notation for the hyperoperation, but I'm going with the presumably most familiar one. I'm sure there are bigger numbers - specifically, Wikipedia mentions Conway chains, where a→b→c = a ↑^c b, and so 2→3→4→5→6→7→8→9→10 should be quite large indeed.

Excellent! I hadn't thought about it, but at a glance I'm pretty sure your ordering and grouping does generate the maximal result. However, the parent comment said as much punctuation as you like, which allows for unbounded expressions with unary operators like the factorial!