To find the maximum value of the expression
$$\frac{1^{a_1} \cdot 2^{a_2} \cdot \ldots \cdot 9^{a_9}}{a_1! \cdot a_2! \cdot \ldots \cdot a_9!}$$
given the constraint $a_1 + a_2 + \ldots + a_9 = 90$, we can use the method of Lagrange multipliers or analyze the properties of the function directly.

1. Setting up the problem:
We need to maximize
$$f(a_1, a_2, \ldots, a_9) = \frac{1^{a_1} \cdot 2^{a_2} \cdot \ldots \cdot 9^{a_9}}{a_1! \cdot a_2! \cdot \ldots \cdot a_9!}$$
subject to the constraint
$$g(a_1, a_2, \ldots, a_9) = a_1 + a_2 + \ldots + a_9 - 90 = 0.$$

2. Using the method of Lagrange multipliers:
We introduce a Lagrange multiplier $\lambda$ and set up the system of equations:
$$\nabla f = \lambda \nabla g.$$
This gives us the equations:
$$\frac{\partial}{\partial a_i} \left( \frac{i^{a_i}}{a_i!} \right) = \lambda \frac{\partial}{\partial a_i} (a_1 + a_2 + \ldots + a_9).$$

3. Simplifying the partial derivatives:
The partial derivative of $f$ with respect to $a_i$ is:
$$\frac{\partial}{\partial a_i} \left( \frac{i^{a_i}}{a_i!} \right) = \frac{i^{a_i} \ln(i) \cdot a_i! - i^{a_i} \cdot a_i \cdot a_i!}{(a_i!)^2} = \frac{i^{a_i} \ln(i) - i^{a_i} \cdot a_i}{a_i!}.$$
Setting this equal to $\lambda$, we get:
$$\frac{i^{a_i} \ln(i) - i^{a_i} \cdot a_i}{a_i!} = \lambda.$$

4. Analyzing the equations:
For each $i$, we have:
$$i^{a_i} \ln(i) - i^{a_i} \cdot a_i = \lambda a_i!.$$
This implies that:
$$\ln(i) - a_i = \frac{\lambda a_i!}{i^{a_i}}.$$

5. Finding the optimal values:
To maximize the expression, we need to balance the terms. By symmetry and the constraint $a_1 + a_2 + \ldots + a_9 = 90$, we can assume $a_i = 2i$ for $i = 1, 2, \ldots, 9$. This satisfies the constraint since:
$$1 \cdot 2 + 2 \cdot 2 + 3 \cdot 2 + \ldots + 9 \cdot 2 = 2(1 + 2 + 3 + \ldots + 9) = 2 \cdot 45 = 90.$$

6. Calculating the maximum value:
Substituting $a_i = 2i$ into the expression, we get:
$$\frac{1^{2} \cdot 2^{4} \cdot 3^{6} \cdot \ldots \cdot 9^{18}}{2! \cdot 4! \cdot 6! \cdot \ldots \cdot 18!}.$$