How is this induction argument here valid I have this proof of 'existence of splitting fields' from Gallian's Contemporary Abstract Algebra (I have seen it elsewhere as well, and it goes in similar lines):

Let $F$ be a field and let $f(x)$ be a non-constant element of $F[x]$. Then there exists a splitting field $E$ for $f(x)$ over $F$.
Proof: We proceed by induction on $\deg f(x)$. If $\deg f(x) = 1$, then $f(x)$ is linear. Now suppose that the statement is true for all fields and all polynomials of degree less than that of $f(x)$. Then, there is an extension $E$ of $F$ in which $f(x)$ has a zero, say, $a_1$ . Then we may write $f(x) = (x - a_1 )g(x)$, where $g(x) \in E[x]$. Since $\deg g(x) < \deg f(x)$, by induction, there is a field $K$ that contains $E$ and all the zeros of $g(x)$, say, $a_2 , . . . , a_n$ . Clearly, then, a splitting field for $f(x)$ over $F$ is $F(a_ 1 , a_ 2 , . . . , a_ n )$.

I can't digest how can we, in the induction step, assume that 'the statement is true for all fields '. I have seen this kind of argument for the first time. Please help me!

Elaborating on Lee Mosher's comment, consider the following assertion, $P(n)$, about a natural number $n$:
$P(n)$: Let $F$ be a field and let $f(x)$ be a non-constant element of $F[x]$ of degree at most $n$. Then there exists a splitting field $E$ for $f(x)$ over $F$.
The statement you want to prove can be written as $(\forall n)(P(n))$ and is thus a candidate for a proof by induction.
The fact that $P(n)$ itself contains two implicit universal quantifiers, including "for all fields $F$", might seem too good to be true ... but mathematics is, quite simply, that good! especially (as we see here) for its ability to package whole infinite families of assertions together in a single statement.
