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.


https://en.wikipedia.org/wiki/Sufficient_statistic#Mathematical_definition
A statistic $t = T(Y)$ is sufficient for underlying parameter $\theta$ precisely if the conditional probability distribution of the data $Y$, given the statistic $t = T(Y)$, does not depend on the parameter $\theta$.

Thus we can describe
$$\overbrace{f(Y=y|T=t,\theta)}^{\text{distribution of the data $X$ given $T = t$, and $\theta$}} = \overbrace{g(Y=y|T=t)}^{\text{function without $\theta$}}$$
Then we can apply that the joint distribution is equal to the conditional distribution multiplied with the distribution of the condition
$$\begin{array}{rcccl}f(Y=y \text{ and } T=t | \theta) &=& f(Y=y|T=t,\theta) &\cdot& f(T=t|\theta) \\ &=& g(Y=y|T=t) &\cdot& f(T=t|\theta)
\\ &=& \dfrac{1}{n^t} \dfrac{t!}{\prod_{i = 1}^n y_i!} &\cdot& f(T=t|\theta)
\end{array}$$
then
$$f(T = t |\theta) = \frac{f(Y=y \text{ and } T=t | \theta) }{\dfrac{1}{n^t}  \dfrac{t!}{\prod_{i = 1}^n y_i!}} $$
and we can use any arbitrary sample $Y$ to compute the above expression on the right side. So let's do it easy and choose $y_1 = t$ while $y_i = 0$ for $i>1$. Then
$$\begin{array}{rcl}f(T = t |\theta) &=& \frac{f(Y_1=t, Y_2 = Y_3 = \dots = Y_n = 0 \text{ and } T=t | \theta) }{\dfrac{1}{n^t}  \dfrac{t!}{t!}} \\
&=& \frac{\lambda^t e^{-\lambda}/t! \cdot \left( \lambda^0 e^{-\lambda}/0! \right)^{n-1}}{\dfrac{1}{n^t}}\\
& =&\frac{\lambda^t e^{-n\lambda}}{t!\dfrac{1}{n^t}} \\
& =&\frac{(n\lambda)^t e^{-n\lambda}}{t!}
\end{array}$$
which is a Poisson distribution with rate parameter $n\lambda$.

Above, I derived the distribution of $T(Y)$ given the conditional distribution of $Y$ given $T(Y)$. I do not get in your question how the quote below is an answer to (b)

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

That answer derives $P(\mathbf{Y} \mid T(\mathbf{Y}))$ instead of $P(T(\mathbf{Y}))$
