[proofplan]
We prove the result by induction on $N$, reducing to the case $N = 2$. For two locally compact Hausdorff spaces $X_1$ and $X_2$, we construct a compact neighbourhood of each point $(x_1, x_2) \in X_1 \times X_2$ by taking the product of compact neighbourhoods from each factor. The Hausdorff property of the product follows from the [Hausdorff and Regularity Preservation Under Products](/theorems/958) theorem. The compactness of a product of two compact sets uses the [Tychonoff Theorem for Finite Products](/theorems/308).
[/proofplan]
[step:Establish the Hausdorff property of the product]
By the [Hausdorff and Regularity Preservation Under Products](/theorems/958) theorem, finite (indeed, arbitrary) products of Hausdorff spaces are Hausdorff in the product topology. Since each $X_i$ is Hausdorff, the product $X_1 \times \cdots \times X_N$ is Hausdorff.
[/step]
[step:Prove local compactness for the product of two factors]
Let $X_1$ and $X_2$ be locally compact Hausdorff spaces. Fix a point $(x_1, x_2) \in X_1 \times X_2$. Since $X_1$ is locally compact, there exists an open set $U_1 \subset X_1$ and a compact set $K_1 \subset X_1$ with $x_1 \in U_1 \subset K_1$. Similarly, there exists an open set $U_2 \subset X_2$ and a compact set $K_2 \subset X_2$ with $x_2 \in U_2 \subset K_2$.
The set $U_1 \times U_2$ is open in $X_1 \times X_2$ (it is a basic open set in the product topology) and contains $(x_1, x_2)$. The set $K_1 \times K_2$ is compact by the [Tychonoff Theorem for Finite Products](/theorems/308), since $K_1$ and $K_2$ are compact. The containment $U_1 \times U_2 \subset K_1 \times K_2$ holds because $U_1 \subset K_1$ and $U_2 \subset K_2$.
Therefore $(x_1, x_2)$ has the open neighbourhood $U_1 \times U_2$ contained in the compact set $K_1 \times K_2$. This shows $X_1 \times X_2$ is locally compact.
[guided]
The key idea is that the product topology has a basis of "rectangles" $U_1 \times U_2$, and compact neighbourhoods in each factor multiply to give a compact neighbourhood of the product point.
Fix $(x_1, x_2) \in X_1 \times X_2$. Local compactness of $X_1$ gives an open $U_1 \ni x_1$ with $U_1 \subset K_1$ for some compact $K_1$. Local compactness of $X_2$ gives an open $U_2 \ni x_2$ with $U_2 \subset K_2$ for some compact $K_2$.
Is $K_1 \times K_2$ compact? By the [Tychonoff Theorem for Finite Products](/theorems/308), the product of two compact spaces is compact in the product topology. Since $K_1$ and $K_2$ are compact (as subspaces of $X_1$ and $X_2$ respectively), $K_1 \times K_2$ is compact.
The product $U_1 \times U_2$ is a basic open set in the product topology, and it satisfies
\begin{align*}
(x_1, x_2) \in U_1 \times U_2 \subset K_1 \times K_2.
\end{align*}
This provides a compact neighbourhood of $(x_1, x_2)$.
[/guided]
[/step]
[step:Extend to $N$ factors by induction]
We proceed by induction on $N$. The base case $N = 1$ is the hypothesis, and $N = 2$ was handled in the previous step.
For the inductive step, assume $X_1 \times \cdots \times X_{N-1}$ is locally compact Hausdorff. The product topology on $X_1 \times \cdots \times X_N$ agrees with the product topology on $(X_1 \times \cdots \times X_{N-1}) \times X_N$ (the product topology is associative for finite products). By the inductive hypothesis, $X_1 \times \cdots \times X_{N-1}$ is locally compact Hausdorff, and $X_N$ is locally compact Hausdorff by assumption. Applying the two-factor case to $(X_1 \times \cdots \times X_{N-1}) \times X_N$, the product is locally compact Hausdorff.
[/step]
