We are given the expression:

$1 + \sqrt{-81} \cdot \sqrt{-9} \cdot \sqrt{-36}$

Step 1: Express Square Roots of Negative Numbers in Terms of $i$

Using the identity $\sqrt{-x} = i\sqrt{x}$, we rewrite each term:

$\sqrt{-81} = i\sqrt{81} = i(9) = 9i$

$\sqrt{-9} = i\sqrt{9} = i(3) = 3i$

$\sqrt{-36} = i\sqrt{36} = i(6) = 6i$

Step 2: Multiply the Terms

Now, compute:

$(9i) \cdot (3i) \cdot (6i)$

First, multiply the first two terms:

$(9i) \cdot (3i) = 9 \cdot 3 \cdot i^2 = 27i^2$

Since $i^2 = -1$, we get:

$27i^2 = 27(-1) = -27$

Now, multiply by $6i$:

$(-27) \cdot (6i) = -162i$

Step 3: Add 1 to the Result

$1 + (-162i) = 1 - 162i$

Final Answer:

$\boxed{1 - 162i}$

Would you like further clarification or additional examples?

Here are five related problems you might find useful:

1. Simplify $\sqrt{-25} \cdot \sqrt{-4}$.
2. Express $(2 + 3i)(4 - i)$ in the form $a + bi$.
3. Find the value of $i^{15}$.
4. Compute $(5i - 2)^2$.
5. Solve for $x$ in $x^2 + 4x + 13 = 0$ over complex numbers.

Tip: Remember that $i^2 = -1$ and use it to simplify expressions with powers of $i$!