Expectation of log-Likelihood

If Y1,Y2,...,Yn has distribution in exponential family depending on the parameters θ1,..., θp and Let L(θ; y) and l(θ; y) denote the likelihood and log-likelihood,

then how can I prove with the help of

Can I consider this as a function of Y because if so then I can take integral of this times with pdf of Y1,Y2..Yn and then I can solve it from there.

But I am not sure if I can consider this as function of Y.

The fact that the score function has a null expectation at the true value of the parameter $$\mathbb E_\theta^\mathbf Y \underbrace{\left[\dfrac{\partial \ell(\theta;\mathbf Y)}{\partial\theta}\right]}_{\substack{\text{random variable}\\ \text{as function of Y}}}=0$$ is unrelated with exponential families. Writing $$\ell(\theta;\mathbf y)=\sum_{i=1}^n \log f(y_i;\theta)\tag{1}$$ and considering as a function of $$\theta$$ the gradient of (1) (in $$\theta$$) is \begin{align}\dfrac{\partial \ell(\theta;\mathbf y)}{\partial\theta} &= \sum_{i=1}^n \dfrac{\partial \log f(y_i;\theta)}{\partial\theta}\\ &=\sum_{i=1}^n \dfrac{1}{f(y_i;\theta)}\dfrac{\partial f(y_i;\theta)}{\partial\theta}\end{align} by standard (calculus) derivation rules. Now, if one transforms the random variable $$Y_i$$ by the mapping $$y \longmapsto \dfrac{1}{f(y;\theta)}\dfrac{\partial f(y;\theta)}{\partial\theta}$$ (which is indexed by $$\theta$$) one obtains another random variable $$\dfrac{1}{f(Y_i;\theta)}\dfrac{\partial f(Y_i;\theta)}{\partial\theta}$$ Then the expectation of this (transformed) random variable under its distribution (indexed by the same $$\theta$$) is \begin{align}\mathbb E_\theta^{Y_i}\left[\dfrac{1}{f(Y_i;\theta)}\dfrac{\partial f(Y_i;\theta)}{\partial\theta}\right]&=\int \dfrac{1}{f(y_i;\theta)}\dfrac{\partial f(y_i;\theta)}{\partial\theta} f(y_i;\theta) \,\text dy_i\\ &=\int \dfrac{\partial f(y_i;\theta)}{\partial\theta} \,\text dy_i\tag{2}\\ &= \dfrac{\partial}{\partial\theta} \int f(y_i;\theta) \,\text dy_i\tag{3}\\ &= \dfrac{\partial}{\partial\theta}\,1\\ &= 0 \end{align} assuming the likelihood $$f(y_i;\theta)$$ is sufficiently regular for (2) and (3) to be equal. (Which means it must be uniformly bounded by an integrable function, following Leibniz's rule.) This proof is actually available on Wikipedia.