Consider the following proof of the fact that $\bar{X}$, the sample mean, is the MLE of parameter $\lambda$ in a Poisson distribution.
Let $x_1, \ldots, x_n$ be the observations of $X_1, \ldots, X_n$ with $X_i \sim \mathcal{P}(\lambda)$. Then we want to solve \begin{align*} \text{argmax}_{\lambda} \mathcal{L}(\lambda \mid x_1, \ldots, x_n) \end{align*} Observe that \begin{align*} P(x_1, \ldots, x_n \mid \lambda) &= \prod_{i=1}^{n} \frac{\lambda^{x_i} e^{-\lambda}}{x_i!} \end{align*} To maximize this with respect to $\lambda$, we observe that \begin{align*} \frac{d}{d\lambda} \ln \left[ \prod_{i=1}^{n} \frac{\lambda^x_i e^{-\lambda}}{x_i!}\right] &= \frac{d}{d\lambda} \sum_{i=1}^{n} \left(\ln \frac{\lambda^{x_i}e^{-\lambda}}{x_i !}\right) \\ &= \frac{1}{\lambda} \sum_{i=1}^{n} x_i - n \end{align*} Then, if we let $S := \sum_{i=1}^{n} x_i$, \begin{align*} \frac{S}{\lambda} - n &= 0 \iff \lambda = \frac{S}{n} \end{align*} where obviously $S/n = \bar{X}$.
My question regards the claim that setting the derivative to zero finds the maximizing $\lambda$. That the log-likelihood has a critical point at $\lambda = \bar{X}$ is proven; however, what guarantees that this critical point is not a minimum?
I am aware that the second derivative test answers this question, because it is negative at $\lambda = \bar{X}$. But the book this comes from is entirely on probability theory and I presume that it's using some fact about the Poisson distribution to implicitly conclude that the point is a maximum. In short, can one conclude that $\lambda = S/n$ is a maximizing point instead of a minimizing point from the nature of the distribution at hand?
The pmf of Poisson distribution is log-concave in $\color{blue} {\lambda}$. This implies that the log of the likelihood function, given by $$ \ln \left[ \prod_{i=1}^{n} \frac{\lambda^{x_i} e^{-\lambda}}{x_i!}\right]= \sum_{i=1}^{n} \left(\ln \frac{\lambda^{x_i}e^{-\lambda}}{x_i !}\right) =\left (\sum_{i=1}^{n} x_i \right )\color{blue} {\ln \lambda} - n\color{blue} {\lambda}-\sum_{i=1}^{n} \ln (x_i !)$$
is concave in $\color{blue} {\lambda}$. Thus, the point that makes its derivative zero is the global optimum.