Given a natural integer n, it can be shown that \(5^{ π β 1} < 3 ^{π + 1}\).
We can conclude from observation that the inequality is true for n = 1, 2, 3, 4, and 5.
Currently, we must determine the smallest integer value of m such that \(3^{ n+1} <2^{ n+m}\) is satisfied.
In the case of n = 1, m's least integer value is 2.
In the case of n = 2, m's least integer value is 3.
In the case of n = 3, m's least integer value is 4.
In the case of n = 4, m's least integer value is 4.
In the case where n = 5, m's least integer value is 5.
Thus, 5 is the lowest integer value of m for which the equation holds true for all values of n.
The correct option is (A): 5.