a finite composition of automorphisms of the same type as given above. Hence, it is an automorphism.

We will prove that $\sigma(f)$ is $X_n$-distinguished of degree $s = \sum_{i=1}^nd_i\nu_i$. First, for all $\mu$ such that $f_\mu(Y)$ is a unit, and $\mu \neq \nu$, we have $\sum_{i=1}^n d_i\mu_i < s$: There exists an index $q$ such that $1 \leq q \leq n$, such that $\mu_1 = \nu_1, \cdots, \mu_{q-1} = \nu_{q-1}$ and $\mu_q < \nu_q$. Therefore $\mu_q \leq \nu_q - 1$ and $$\begin{aligned}
 \sum_{i=1}^nd_i\mu_i \leq \sum_{i=1}^{q-1}d_i\nu_i + d_q(\nu_q-1) + t\sum_{i=q+1}^nd_i =
 \sum_{i=1}^qd_i\nu_i - 1 < \sum_{i=1}^nd_i\nu_i = s.\end{aligned}$$ Now, $$\begin{aligned}
 \sigma(f) &=& \sum_\mu f_\mu(Y)(X_1 + X_n^{d_1})^{\mu_1}\cdots(X_{n-1} + X_n^{d_{n-1}})^{\mu_{n-1}}X_n^{\mu_n} \\
&\equiv& \sum_{\substack{\mu\\  f_{\mu}(Y) \notin (Y)}}f_{\mu}(Y)\sum_{\substack{\lambda_1, \cdots, \lambda_{n-1} \\ 0\leq \lambda_i \leq \mu_i}}\binom{\mu_1}{\lambda_1}\cdots\binom{\mu_{n-1}}{\lambda_{n-1}}X_1^{\mu_1 - \lambda_1}\cdots X_{n-1}^{\mu_{n-1} - \lambda_{n-1}}X_n^{d_1\lambda_1 + \cdots + d_{n-1}\lambda_{n-1} + \mu_n} \\
&\equiv& \sum g_iX^i_n \mod(Y)\end{aligned}$$ where the $g_i$ are elements of $F[X_1, \cdots, X_{n-1}]$. Therefore, $\sigma(f) \mod (Y)$ is a polynomial in $X_n$ of degree less than or equal to $s$, and $X_n^{d_1\lambda_1 + \cdots + d_{n-1}\lambda_{n-1} + \mu_n} = X_n^s$ if, and only if, $\mu_n = \nu_n$ and $\lambda_i = \mu_i = \nu_i$ for $i = 1, \cdots, n-1$. Thus, we have $g_s = f_\nu(Y) \mod(Y)$, but $f_\nu(Y)$ 