Let $N_t:=\mathbb{I}_{(t\leq \tau \leq T)}$ a counting process with Poisson distribution. I know that $X\sim \mathrm{Po} \Rightarrow P(X=k):=\frac{e^{-\lambda}\lambda^{k}}{k!}$.
Assuming that $\lambda:=\int_{t}^{s}\lambda_udu$.
Why $\mathbb{E}^{\mathbb{Q}}[e^{-\int_{t}^{\tau}r_sds}\mathbb{I}_{(t\leq \tau\leq T)}]=\mathbb{E}^{\mathbb{Q}}[\int_{t}^{T}e^{-\int_{t}^{\tau}r_sds}\lambda(s)e^{-\int_{0}^{s}\lambda_udu}]$?
I've only understood (trivially) that $\mathbb{E}[X]=\int_{\mathbb{R}}xf(x)dx\rightarrow \mathbb{E}^{\mathbb{Q}}[\int_{t}^{T}...]$.