How does the result $\dfrac{1}{n^T} \dfrac{T!}{\prod_{i = 1}^n Y_i!}$ tell us what distribution $T(\mathbf{Y})$ is? This follows on from my question here.
I have the following problem:

Let $Y_1, \dots, Y_n$ be a random sample from a Poisson distribution $\text{Pois}(\lambda)$. Recall, the $\text{Pois}(\lambda)$ distribution has the probability function $f_{\lambda}(y) = e^{-\lambda} \dfrac{\lambda^y}{y!}$, if $y = 0, 1, 2, 3, \dots$, and $\lambda > 0$.
(a) Show that $T(\mathbf{Y}) = \sum_{i = 1}^n Y_i$ is a sufficient statistic for $\lambda$ using the Fisher-Neyman factorisation theorem.
(b) What is the distribution of $T(\mathbf{Y})$? Obtain this result directly using the definition of a sufficient statistic.

For (a), we have that $L(\lambda, \mathbf{y}) = \prod_{i = 1}^n e^{-\lambda}\dfrac{\lambda^{y_i}}{y_i!} = e^{-n \lambda} \dfrac{\lambda^{\sum_{i = 1}^n y_i}}{\prod_{i = 1}^n y_i!}$. So $T(\mathbf{y}) = \sum_{i = 1}^n y_i$, $g(t, \lambda) = e^{-n \lambda} \lambda^t$ and $h(\mathbf{y}) = \dfrac{1}{\prod_{i = 1}^n y_i!}$.
For (b), the solution is given as follows:

$$T(\mathbf{Y}) \sim \text{Pois}(n \lambda)$$
$$P(\mathbf{Y} \mid T(\mathbf{Y})) = \dfrac{P(\mathbf{Y}, T(\mathbf{Y}))}{P(T(\mathbf{Y}))} = \dfrac{\prod_{i = 1}^n e^{-\lambda} \dfrac{\lambda^{Y_i}}{Y_i!}}{e^{-n \lambda}\dfrac{n^T \lambda^T}{T!}} = \dfrac{1}{n^T} \dfrac{T!}{\prod_{i = 1}^n Y_i!}$$

How does the result $\dfrac{1}{n^T} \dfrac{T!}{\prod_{i = 1}^n Y_i!}$ tell us what distribution $T(\mathbf{Y})$ is? Looking at it, I don't understand what point the problem was trying to make. Yes, we derived this result using the definition of sufficient statistic, but $\dfrac{1}{n^T} \dfrac{T!}{\prod_{i = 1}^n Y_i!}$ seems like quite an ugly result, in that it does not seem very/immediately insightful as to what the distribution of $T(\mathbf{Y})$ is.

Here is a slide from my mathematical statistics course that can help:

