How do we prove that if a < b, then a^n < b^n?

only true if a>0 and b>0, or b>0 and n is an integer not divisible by 2 (i.e., if a<O, a^n is negative).

If a and b are both greater than 0 and n is a positive integer, then when a<b, (a/b)^n must be less than one because a/b must be less than 1. thus, a^n must be less than 1*b^n.

-2 < 0
(-2)^2 < 0
4 < 0. False.
Disproven.

3 < 4
3^(-2) < 4^(-2)
1/9 < 1/16. False
Disproven.

This is not true if, for example,
a = -2, b = -3, and n = 2.

How about induction, assuming n is a positive integer?
Let n = 1.
a^1 = a < b = b^1
Therefore, the hypothesis is true for n = 1.
Suppose the hypothesis is true for n and consider n + 1.
a^n < b^n
a^n * a < b^n * b
a^(n + 1) < b^(n + 1)
Therefore, the hypothesis being true for n implies it's true for n + 1.
Therefore, the hypothesis is proved by induction.

∵a<b
∴（a/b)<1
∴(a^n/b^n)=(a/b)^n<1, (if n>0)
∴a^n<b^n

update:
a>0,b>0
thx @Pope pointed out it.

All you need to disprove this is one case, use a negative n.
FALSE
So this cannot be proven.

As you will see from many of these answers, you have not defined a complete question. You have to START by saying what a, b, and n represent, otherwise is it easy to make a statement such as "if a < b, then a^n < b^n?" mean just about anything. Your statement is true only if a, b, and n are positive integers. But you did not say that, so the problem is ill-defined and therefore unanswerable.

If a < b, then a^n < b^n
Let a = 1, b = 2 and c = 3
1 < 2 and 1^3 < 2^3