Step 1: Simplify the complex fraction by rationalizing the denominator.

$\frac{1+i}{1-i} = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{(1-i)(1+i)}$

Step 2: Expand numerator and denominator.

Numerator: $(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$

Denominator: $(1-i)(1+i) = 1 - i^2 = 1 - (-1) = 2$

Step 3: Simplify the fraction.

$\frac{1+i}{1-i} = \frac{2i}{2} = i$

Step 4: Substitute back into the given equation.

$\left( \frac{1+i}{1-i} \right)^m = i^m = 1$

Step 5: Apply the property of powers of i from the context.

From the context: $i^4 = (i^2)^2 = (-1)^2 = 1$

Also: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$

Therefore, the least positive integral value of $m$ for which $i^m = 1$ is m = 4.