WTP: If $P\in {\Bbb C[x]}$ has a leading coefficient $a_n$, then P factorises completely into linear factors in $\Bbb C$, $$P(x) = a_n(x - \alpha)(x - \alpha_1)(x - \alpha_2)\ldots(x - \alpha_n)$$
I have been told to use induction on $n$ (ignored base case here). Firstly I let the degree of P be $n$ ($n > 0$) and assumed that the result is true for polynomials of degree at most $n-1$. We know that P has at least one root in $\Bbb C$ due to Fundamental Theorem of Algebra. Let us call that root $\alpha$ which then implies that $(x-\alpha)$ divides $P$. We then rewrite $P$ in the form of $P = (x-\alpha)Q$ where the degree of $Q$ is $n-1$. Then by the inductive assumption, we know that $Q$ can be expressed in the form shown in the statement above i.e
$$ Q(x) = a_n(x - \alpha)(x - \alpha_1)(x - \alpha_2)\ldots(x - \alpha_n)$$
As $P = (x-\alpha)Q$, we can see that $P$ can be written in the form required.