Let us expand the term \(((11^{12})^{13})^{14}\)
\(((11^{12})^{13})^{14}\)= \(((9+2)^{12})^{13})^{14}\)
When we expand the term using binomial expansion, all terms except the last will be divisible by 9
\(((2^{12})^{13})^{14}\)
\(((2^3)^4)^{13})^{14}\)
\(((8)^4)^{13})^{14}\)
\(((9-1)^4)^{13})^{14}\)
Again only last term will not be divisible by 9, when the term is expanded
\(((-1)^4)^{13})^{14}\)
\((1^{13})^{14}=1\)
So remainder =1.

We can also find pattern
11 leaves a remainder 2
11^2.....2^2
11^3.....2^3=8
11^4......2^4=16=9+7
11^5......2^5=32=27+5
11^6......2^6=64=63+1
11^7......2^7=128=126+2
So pattern repeats after every 6 terms => 2,4,8,7,5,1,2,4,8....
Now the power 12 is multiple of 6, so 11^12 will leave 1 as the remainder.

A