天文數字問題大到你估唔到

1^n = 10...0 have n-1 zero.


so... (1) ... have 1兆個0
n = 1兆 + 1


(2) 1兆 , n = 1兆 - 1
1兆^1兆 = n^2

.....if n is a positive integer (it is 1), n^2 > n -1.... therefore (2) is bigger

10^n = 1...0 <==n zero
let k = 1兆
1000......0000
|__________|
1000....0000個0
|_________|
1000....0000個0
...
|__________|
1000....0000個0
|________|
1兆個0

= 10^(10^(10^(....(10^(10^k))....)))
it has (10^(10^(....(10^(10^k))....))) zero
2 1兆^1兆^1兆......^1兆^1兆
= k^k^k......^k^k
if k^k^k......^k^k = ((...((k^k)^k)...^k)^k)
= k^(k^2)
it has (k^2)log(k,10) < 1's answer as 10^k is already larger than k^2

if k^k^k......^k^k = k^(k^(k(....^(k^(k^k))....)))
> 1's answer as k >1

So no concludion can be drawn if (2) is not stated clearly