Show that there exists a unique measure $\pi$ on $\mathcal{B}[0,\infty)\otimes\mathcal{P}(\mathbb{N})$

Question

I have already shown that $\Gamma\in \mathcal{B}[0,\infty)\otimes\mathcal{P}(\mathbb{N})$ iff $\Gamma = \bigcup_{j\in\mathbb{N}} A_j\times \{j\}$ where $A_j \in\mathcal{B}[0,\infty)$ for all $j$. I am now trying to show existence and uniqueness of the measure defined as follows $$ \pi(B\times\{n\}) = \int_B e^{-x}\frac{x^n}{n!}\mu(dx)$$ where $\mu$ is a bounded measure on $\mathcal{B}[0,\infty)$. I struggle with the third criterion to show that $\pi$ is a measure. Clearly $\pi(\emptyset) = 0$ and $\mu(A)\geq 0$ for any $A\in \mathcal{B}[0,\infty)\otimes\mathcal{P}(\mathbb{N})$ since the integrand in is strictly positive. The most natural way to extend this function to disjoint unions is to have if $(A\times\{i\}) \cap (B\times\{j\})=\emptyset$ where $A,B\in\mathcal{B}[0,\infty)$, then $$ \pi((A\times\{i\}) \cup (B\times\{j\})) = \int_A e^{-x}\frac{x^i}{i!}\mu(dx) + \int_B e^{-x}\frac{x^j}{j!}\mu(dx)$$ This extension immediately implies that for any $\Gamma\in \mathcal{B}[0,\infty)\otimes\mathcal{P}(\mathbb{N})$ we have $$ \pi(\Gamma) = \sum_{n=1}^\infty \int_{A_n} e^{-x}\frac{x^n}{n!}\mu(dx)$$ Let $\Gamma_n$ be any disjoint sequence of sets in our $\sigma-$algebra. Then we may write $\bigcup \Gamma_n = \bigcup_{n,i} ( A_{n,i}\times\{n\})$ where $A_{n,i}\cap A_{n,j}=\emptyset$ if $i\not=j$.Hence $$ \begin{align*}\pi\Big(\bigcup_n \Gamma_n\Big) &= \sum_{n=1}^\infty \int_{\bigcup_i A_{n,i}}e^{-x}\frac{x^{n}}{n!}\mu(dx) \\ &= \sum_{n=1}^\infty \sum_{i=1}^\infty \int_{\bigcup_i A_{n,i}}e^{-x}\frac{x^{n}}{n!}\mu(dx) = \sum_{n=1}^\infty \mu(\Gamma_n)\end{align*}$$ I believe this should show that $\pi$ is a measure. I can show that our space is finite using that $\sum_{k=1}^n e^{-x}\frac{x^k}{k!}$ is dominated by $g=1$ on $([0,\infty), \mathcal{B}[0,\infty),\mu)$ to get $$\begin{align*}\pi([0,\infty)\times\mathbb{N}) &= \sum_{n=1}^\infty\int_{[0,\infty)} e^{-x}\frac{x^n}{n!}\mu(dx)\\ &= \int_{[0,\infty)}e^{-x}\sum_{n=1}^\infty \frac{x^n}{n!}\\ &= \int_{[0,\infty)} \mu(dx) = \mu([0,\infty))< \infty\end{align*}$$ So our product space is finite. But I am not sure how to use this fact to show that $\pi$ is unique. I know I probably need to use the exhaustive sequence $[0,n)\times\{1,\ldots n\}$ but I cannot figure out how.