= Solution
Use these standard <ultrafilter> facts: a space is <compact> if and only if every <ultrafilter> on it converges; the image <ultrafilter> under a map is an <ultrafilter>; and a convergent <ultrafilter> contains every <neighborhood> of its limit.
Let $X=\prod_{\lambda\in\Lambda}X_\lambda$ with all factors <compact>. An empty factor makes $X$ empty and hence <compact>. Otherwise, given an <ultrafilter> $\mathcal U$ on $X$, push it forward by each <product projection> to obtain an <ultrafilter> $\mathcal U_\lambda$ on $X_\lambda$. Choose a limit $x_\lambda$ of each, and put $x=(x_\lambda)_\lambda$. A basic product <neighborhood> of $x$ is an intersection of finitely many inverse images of coordinate <neighborhoods>. Each inverse image belongs to $\mathcal U$, so their <finite intersection> does too. Thus $\mathcal U$ converges to $x$, proving \b[<compactness> of the product]. Uniqueness of coordinate limits is not needed, so the factors need not be <Hausdorff>.
For <local compactness>, suppose only the factors indexed by a finite set $F$ may be noncompact. At $x\in X$ choose, for each $\lambda\in F$, a <compact> <neighborhood> $K_\lambda$ of $x_\lambda$, with an <open neighborhood> $U_\lambda\subseteq K_\lambda$. For other indices put $K_\lambda=X_\lambda$. The <compact> product $\prod_\lambda K_\lambda$ contains the open product <neighborhood> restricting just the finitely many coordinates in $F$ to $U_\lambda$. Hence every <point> has a <compact> <neighborhood> and \b[the product is locally <compact>].
The unrestricted assertion is \b[false]. Each factor of $\mathbb R^{\mathbb N}$ is locally <compact>, but its product is not. Any basic <neighborhood> restricts only finitely many coordinates. Choose a free coordinate $j$; its projection of that <neighborhood> is all of $\mathbb R$. A <compact> <neighborhood> containing it would consequently have <compact> image equal to $\mathbb R$, impossible. This shows directly that no <point> has a <compact> <neighborhood>.
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