\omega F[X_1, \cdots, X_{n-1}; Y, \lambda][X_n]$, trivially.

$F[X; Y, \lambda] = F[X_1, \cdots, X_n; Y, \lambda]$ is factorial, for all $n \geq 1$.

First assume that $n=1$. Suppose that $f$ is an element of $F[X; Y, \lambda]$. Write $f = e\cdot Y^d  \omega $, where $\omega$ is a unitary polynomial in $X$ of degree $s$ in $F[[Y]][X]$, and $e$ is a unit in $F[X; Y, \lambda]$. We can factor $\omega = uq_1\cdots q_m$ into irreducible factors and a unit in $F[[Y]][X]$ because this ring is factorial. We want to show that these factors are still irreducible in $F[X; Y, \lambda]$. Suppose that $q_i$ is not irreducible modulo $\omega F[[Y]][X]$, then $q_i \equiv ab \mod \omega$, so there exists $g \neq 0$ such that $q_i = ab + g\omega$. However, by the uniqueness of the division algorithm $g = 0$, thus, $a$ or $b$ is a unit modulo $\omega$. Therefore, $q_i$ is irreducible in $F[[Y]][X]/\omega F[[Y]][X] \simeq F[X; Y, \lambda]/\omega F[X; Y, \lambda]$.

If $q_i$ is not irreducible in $F[X; Y, \lambda]$, then there exists elements $a$ and $b$ in $F[X; Y, \lambda]$, such that $q_i = ab$. Without loss of generality, $b$ must be a unit modulo $\omega$, so $b 