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To understand this type of computation, you should write down the details explicitly. To lighten the notation, I set $a=z/\lambda^3$ and $b_l=U_l/l!$. Then, writing down explicitly the constraint as an indicator function, \begin{align} \sum_N \sum_{\{m_l\}} 1_{\{\sum_l lm_l = N\}} a^N \prod_{l}\frac{1}{m_l!} b_l^{m_l} &= \sum_N \sum_{\{m_l\}} 1_{\{\sum_l lm_l = N\}} \prod_{l} \Bigl( a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \Bigr)\\ &= \sum_{\{m_l\}} \prod_{l} \Bigl( a^{lm_l} \frac{1}{m_l!} b_l^{m_l}\Bigr) \sum_N 1_{\{\sum_l lm_l = N\}} \\ &= \sum_{\{m_l\}} \prod_{l} a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \\ &= \prod_l \sum_{m_l} a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \\ &= \prod_l \sum_{m_l} \frac{1}{m_l!} (a^lb_l)^{m_l} \\ &= \prod_l \exp( a^lb_l ). \end{align} For the first identity, I replaced $N$ by $\sum_l lm_l$ (I can then write $a^N$ as a product over $l$). The third identity follows from the fact that exactly one term in the sum over $N$ is equal to $\sum_l lm_l$. The fourth identity is a consequence of the fact that the summand is completely factorized over $l$, so that $m_l$ can be summed separately for each $l$. The last identity is just the Taylor series of the exponential.

To understand this type of computation, you should write down the details explicitly. To lighten the notation, I set $a=z/\lambda^3$ and $b_l=U_l/l!$. Then, writing down explicitly the constraint as an indicator function (so that the sum over $\{m_l\}$ is now unrestricted), \begin{align} \sum_N \sum_{\{m_l\}} 1_{\{\sum_l lm_l = N\}} a^N \prod_{l}\frac{1}{m_l!} b_l^{m_l} &= \sum_N \sum_{\{m_l\}} 1_{\{\sum_l lm_l = N\}} \prod_{l} \Bigl( a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \Bigr)\\ &= \sum_{\{m_l\}} \prod_{l} \Bigl( a^{lm_l} \frac{1}{m_l!} b_l^{m_l}\Bigr) \sum_N 1_{\{\sum_l lm_l = N\}} \\ &= \sum_{\{m_l\}} \prod_{l} a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \\ &= \prod_l \sum_{m_l} a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \\ &= \prod_l \sum_{m_l} \frac{1}{m_l!} (a^lb_l)^{m_l} \\ &= \prod_l \exp( a^lb_l ). \end{align} For the first identity, I replaced $N$ by $\sum_l lm_l$ (I can then write $a^N$ as a product over $l$). The third identity follows from the fact that exactly one term in the sum over $N$ is equal to $\sum_l lm_l$. The fourth identity is a consequence of the fact that the summand is completely factorized over $l$, so that $m_l$ can be summed separately for each $l$. The last identity is just the Taylor series of the exponential.

(Note: there's a typo in the last line of your computation.)

To understand this type of computation, you should write down the details explicitly. To lighten the notation, I set $a=z/\lambda^3$ and $b_l=U_l/l!$. Then, writing down explicitly the constraint as an indicator function, \begin{align} \sum_N \sum_{\{m_l\}} 1_{\{\sum_l lm_l = N\}} a^N \prod_{l}\frac{1}{m_l!} b_l^{m_l} &= \sum_N \sum_{\{m_l\}} 1_{\{\sum_l lm_l = N\}} \prod_{l} a^{lm_l} \frac{1}{m_l!} b_l^{m_l}\\ &= \sum_{\{m_l\}} \prod_{l} a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \sum_N 1_{\{\sum_l lm_l = N\}} \\ &= \sum_{\{m_l\}} \prod_{l} a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \\ &= \prod_l \sum_{m_l} a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \\ &= \prod_l \sum_{m_l} \frac{1}{m_l!} (a^lb_l)^{m_l} \\ &= \prod_l \exp( a^lb_l ). \end{align} For the first identity, I replaced $N$ by $\sum_l lm_l$ (I can then write $a^N$ as a product over $l$). The third identity follows from the fact that exactly one term in the sum over $N$ is equal to $\sum_l lm_l$. The fourth identity is a consequence of the fact that the summand is completely factorized over $l$, so that $m_l$ can be summed separately for each $l$. The last identity is just the Taylor series of the exponential.

To understand this type of computation, you should write down the details explicitly. To lighten the notation, I set $a=z/\lambda^3$ and $b_l=U_l/l!$. Then, writing down explicitly the constraint as an indicator function, \begin{align} \sum_N \sum_{\{m_l\}} 1_{\{\sum_l lm_l = N\}} a^N \prod_{l}\frac{1}{m_l!} b_l^{m_l} &= \sum_N \sum_{\{m_l\}} 1_{\{\sum_l lm_l = N\}} \prod_{l} \Bigl( a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \Bigr)\\ &= \sum_{\{m_l\}} \prod_{l} \Bigl( a^{lm_l} \frac{1}{m_l!} b_l^{m_l}\Bigr) \sum_N 1_{\{\sum_l lm_l = N\}} \\ &= \sum_{\{m_l\}} \prod_{l} a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \\ &= \prod_l \sum_{m_l} a^{lm_l} \frac{1}{m_l!} b_l^{m_l} \\ &= \prod_l \sum_{m_l} \frac{1}{m_l!} (a^lb_l)^{m_l} \\ &= \prod_l \exp( a^lb_l ). \end{align} For the first identity, I replaced $N$ by $\sum_l lm_l$ (I can then write $a^N$ as a product over $l$). The third identity follows from the fact that exactly one term in the sum over $N$ is equal to $\sum_l lm_l$. The fourth identity is a consequence of the fact that the summand is completely factorized over $l$, so that $m_l$ can be summed separately for each $l$. The last identity is just the Taylor series of the exponential.