Write the inhomogeneous Poisson point process on a Borel subset $A$ of $\mathbb R^d$ as $$\sum_{i=1}^N\delta_{X_i},$$ where $\delta_a$ is the Dirac probability measure at point $a$, $N$ is the random variable whose value is the number of points that appeared, and the $X_i$'s are the random locations of those points. Then, for each natural $n$, the conditional density $p_n$ of $(X_1,\dots,X_n)$ given that $N=n$ is given by the formula $$p_n(x_1,\dots,x_n)=\Big(\int_A\Lambda(x)\,dx\Big)^{-n}\,\prod_{i=1}^n \Lambda(x_i)\tag1$$ for $(x_1,\dots,x_n)\in A$, provided that $\int_A\Lambda(x)\,dx>0$. So, your likelihood function is this conditional density function $p_n$.

Formula (1) is intuitively obvious, because, at least when $\Lambda$ is continuous and $A$ open, the probability that there will be a point of the Poisson point process in a small neighborhood $U$ of a point $x\in A$ will be approximately $\Lambda(x)|U|$, where $|U|$ is the Lebesque measure of $U$.

The normalizing factor $\Big(\int_A\Lambda(x)\,dx\Big)^{-n}$ is what makes $p_n$ a probability density. However, for the usual inferences involving the likelihood function (say in the maximum likelihood estimation), the choice of a normalizing factor does not matter.

Up to such an inessential normalizing factor, the answer you linked (given for the case when the dimension is $d=1$) is the same as your likelihood. However, the phrase "distribution of the $n$ first events", used in that answer, does not make sense to me: distribution of [...] events?

Write the inhomogeneous Poisson point process on a Borel subset $A$ of $\mathbb R^d$ as $$\sum_{i=1}^N\delta_{X_i},$$ where $\delta_a$ is the Dirac probability measure at point $a$, $N$ is the random variable whose value is the number of points that appeared, and the $X_i$'s are the random locations of those points. Then, for each natural $n$, the conditional density $p_n$ of $(X_1,\dots,X_n)$ given that $N=n$ is given by the formula $$p_n(x_1,\dots,x_n)=\Big(\int_A\Lambda(x)\,dx\Big)^{-n}\,\prod_{i=1}^n \Lambda(x_i)\tag1$$ for $(x_1,\dots,x_n)\in A$, provided that $\int_A\Lambda(x)\,dx>0$. So, your likelihood function is this conditional density function $p_n$.

Formula (1) is intuitively obvious, because, at least when $\Lambda$ is continuous and $A$ open, the probability that there will be a point of the Poisson point process in a small neighborhood $U$ of a point $x\in A$ will be approximately $\Lambda(x)|U|$, where $|U|$ is the Lebesgue measure of $U$.

The normalizing factor $\Big(\int_A\Lambda(x)\,dx\Big)^{-n}$ is what makes $p_n$ a probability density. However, for the usual inferences involving the likelihood function (say in the maximum likelihood estimation), the choice of a normalizing factor does not matter.

Up to such an inessential normalizing factor, the answer you linked (given for the case when the dimension is $d=1$) is the same as your likelihood. However, the phrase "distribution of the $n$ first events", used in that answer, does not make sense to me: distribution of [...] events?