Step 1: Recall the auxiliary circle of a hyperbola.
For the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, the auxiliary circle is defined as $x^2 + y^2 = a^2$. It is a circle centred at the origin with radius equal to the semi-transverse axis $a$.
Step 2: Extract $a^2$ from the given auxiliary circle.
The given auxiliary circle is $x^2 + y^2 = 16$. Comparing: \[a^2 = 16 \implies a = 4\]
Step 3: Use the given point to find $b^2$.
The point $(4\sqrt{2},\ 3)$ lies on the hyperbola. Substituting $x = 4\sqrt{2}$, $y = 3$, $a^2 = 16$: \[\frac{(4\sqrt{2})^2}{16} - \frac{3^2}{b^2} = 1\] \[\frac{32}{16} - \frac{9}{b^2} = 1\] \[2 - \frac{9}{b^2} = 1\]
Step 4: Solve for $b^2$.
\[\frac{9}{b^2} = 1 \implies b^2 = 9\]
Step 5: Apply the eccentricity formula.
For a hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$: \[e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}\]
Step 6: Verify $e > 1$.
Since $e = \dfrac{5}{4} > 1$, this is consistent with a hyperbola (eccentricity is always greater than 1 for a hyperbola).
Step 7: State the final answer.
\[ \boxed{\frac{5}{4}} \]