Some problems about the proof of a theorem There's a theorem in my book (Complexity and cryptography by Talbot and Welsh, chapter 4) where I don't understand some parts of its proof:

THEOREM: Suppose $f \in \mathbb Z[x_1,..., x_n]$ has degree at most $k$ and is
  not identically zero. If $a_1,...,a_n$ are chosen independently and
  uniformly at random from $\{1,...,N\}$then $$Pr[f(a_1,...,a_n) = 0]
 \leq \frac kN$$ Proof: We use induction on n.$\color{blue}{\text{For  }
 n = 1}$ the result holds since a polynomial of degree at most $k$ in a
  single variable has at most $k$ roots. So let $n > 1$ and write $$f =
 f_0+ f_1x_1+ f_2x_1^2+\dots+ f_tx_1^t$$ where $f_0,\dots, f_t$ are
  polynomials in $x_2, x_3,... x_n$; $f_t$ is not identically zero and
   $t \geq 0$. If $t = 0$ then $f$ is a polynomial in $n −1$ variables so
  the result holds. So we may suppose that $1 \leq t \leq k$ and $f_t$
  is of degree at most $k −t$.   We let $E_1$ denote the event
  ‘$f(a_1,\dots,a_n) = 0$’ and $E_2$ denote the event
   ‘$f_t(a_2,\dots,a_n) = 0$’. Now $$Pr[E_1] = Pr[E_1 | E_2]
 Pr[E_2]+Pr[E_1 | not E_2] Pr[not E_2] \leq Pr[E_2]+Pr[E_1 | not
 E_2]\tag{1}$$ Our inductive hypothesis implies that $$Pr[E_2] =
 Pr[f_t(a_2,...,a_n) = 0] \leq\frac{k −t}N$$ since $f_t$ has degree at
  most $k −t$. Also $$Pr[E_1 | not E_2] \leq \frac tN$$ This is true
  because a_1 is chosen independently of $a_2,...,a_n$, so if
  $a_2,...,a_n$ are ﬁxed and we know that $f_t(a_2,...a_n) = 0$ then f
  is a polynomial in $x_1$ that is not identically zero. Hence f , as a
  polynomial in $x_1$, has degree t and so has at most t roots. Putting
  this together we obtain $$Pr[f(a_1,...,a_n) = 0] \leq \frac{k −t}N
 +\frac tN \leq \frac kN$$ as required.



*

*I don't understand how it holds for $n=1$ and $t=0$.

*Where did the formula $(1)$ come from? I don't get it. 

 A: For your first question, notice that the given proof doesn't actually have a case "$n=1$ and $t=0$", because the case $n=1$ is treated separately, and $t$ is introduced only in the argument for $n>1$.  If you nevertheless want to consider $n=1$ and $t=0$, then you're dealing with a constant polynomial (because $t=0$) of one variable (because $n=1$).  Furthermore, the constant isn't $0$, because of the hypothesis in the theorem that $f$ isn't identically zero.  So $f$ is a non-zero constant, and the probability that $f(a_1,\dots,a_n)=0$ is zero, as the theorem claims.
For your second question, equation (1) comes from combining three facts.  First, the event $E_1$ is the disjoint union of the two events $E_1\cap E_2$ and $E_1\cap\sim E_2$ (where I'm using $\sim$ for "not").  Second, by definition of conditional probability, 
$$
Pr[E_1\cap E_2]= Pr[E_1\mid E_2]\cdot Pr[E_2],
$$
and similarly with $\sim E_2$ in place of $E_2$.  That gives the equality at the beginning of (1).  Third, all probabilities and all conditional probabilities are between $0$ and $1$, inclusive.  Applying that to $Pr[E_1\mid E_2]$ and to $Pr[\sim E_2]$, you get the inequality at the end of (1).
EDIT, in answer to a comment: The case $n=1$ is covered by the fact from elementary algebra that a polynomial, in one variable, of degree $k$ cannot have more than $k$ roots.  So of the $N$ possible choices for $a_1$, at most $k$ make $f(a_1)=0$.  So, if $a_1$ is chosen art random from the $N$ possibilities, the probability of getting $f(a_1)=0$ is at most $k/N$.
For $n>1$ and $t=0$, the polynomial $f$, though apparently a polynomial in $n$ variables $x_1$ to $x_n$, doesn't actually involve $x_1$ at all; that's what $t=0$ means.  So it's really a polynomial in $n-1$ (or fewer) variables.  Since the proof is by induction on $n$, the desired conclusion for this $f$ will be given by the induction hypothesis.
