There are infinitely many choices of $(\alpha_1,\dots,\alpha_n)$ such that $f(\alpha_1,\dots,\alpha_n)\neq 0$ I'm trying to solve this exercise in the page 10 of this book

Maybe I'm forgetting something, but I couldn't solve this exercise, I need a hint or something to begin to solve this question.
Thanks in advance
 A: Deal  with the case $n = 1$ first. Here you use that a non-zero polynomial in one variable over a field has only finitely many roots.
When $n > 1$, argue by induction.  

(Thanks to user78535 for suggesting some extra care here.)

If $X_1$ does not appear in $f$, then we have a polynomial in $n-1$ variables, and induction applies. 
If $X_1$ does appear in $f$, write $f$ as a polynomial in $X_1$, with coefficients in $K[X_2, \dots, X_n]$. Clearly the coefficient of one of the powers $X_1^i$, for $i > 0$ has to be non-zero as a polynomial. Thus by induction there are infinitely many choices of $(\alpha_{2}, \dots, \alpha_{n}) \in (K \setminus \Lambda)^{n-1}$ for which this coefficient is non-zero as an element of $K$, when evaluated at $(\alpha_{2}, \dots, \alpha_{n})$. 
Now appealing to the case $n = 1$, for each such choice of $(\alpha_{2}, \dots, \alpha_{n})$ there are infinitely many choices of $\alpha_{1}$ such that $f(\alpha_{1}, \dots, \alpha_{n}) \ne 0$.
A: Hint: the case $\,n=1\,$ is clear, by a polynomial $\neq 0\,$ over a field has no more roots than its degree. For $\,n>1,\,$ write $f = c_k\, x_n^k + \cdots + c_0,\ c_i \in K[x_1,\ldots,x_{n-1}],\ c_k\neq 0.\,$ By induction we can specialize $\,(x_1,\ldots,x_{n-1}) = (a_1,\ldots,a_{n-1})\in (K \setminus \Lambda)^{n-1}$ so  the leading coeff $\,c_k\,$ stays $\ne 0,\,$ yielding $\, 0\ne g(x_n) = f(a_1,\ldots,a_{n-1},x_n) \in K[x_n].\,$ By induction there are infinitely many $\, a_n \in K \setminus \Lambda\,$ such that $\,g(a_n) = f(a_1,\ldots,a_n) \ne 0.\ \ $ QED
