We are define the Ramanujan $\tau$-function as
$\tau(n):=$ is the Fourier coefficients of $\displaystyle{\Delta ( z ) = q \prod\limits_{n=1}^{\infty} ( 1 - q^{n} )^{24}}$. \begin{equation*} \Delta ( z ) = q \prod\limits_{n=1}^{\infty} ( 1 - q^{n} )^{24} = \sum\limits_{n=1}^{\infty} \tau ( n ) q^{n} \text{ where } q = e^{2 \pi i z} \end{equation*}
We want to prove that: $\displaystyle{| \tau ( n ) | \neq 1}$ for all $n \geq 2$.
We assume by contradiction that: there is a positive integer $n \geq 2$ such that $\displaystyle{| \tau ( n ) | = 1}$. We know that: $\tau$ is multiplicative function! Let $\displaystyle{n=p_{1}^{a_{1}} p_{2}^{a_{2}} \ldots p_{k}^{a_{k}}}$ where $p_{i}$ are different prime numbers and $a_{i} \geq 1$. We have \begin{equation*} 1 = | \tau ( n ) | = | \tau ( p_{1}^{a_{1}} ) | \text{ } | \tau ( p_{2}^{a_{2}} ) | \ldots | \tau ( p_{k}^{a_{k}} ) | \Longrightarrow \exists i \in \{ 1 , 2 , \ldots k \} : | \tau ( p_{i}^{a_{i}} ) | = 1. \end{equation*} And we know that: $\displaystyle{\tau ( p^{m+1} ) = \tau ( p ) \tau ( p^{m} ) - p^{11} \tau ( p^{m-1} ) }$ for all $m \geqslant 1$ and for every prime number $p$.
So if we want to verify that if there exists a positive integer number $n\geq 2$ such that $| \tau ( n ) | = 1$, it is enough to check if the numbers $\displaystyle{\pm 1}$ appear in the sequence $\tau ( p ) , \tau ( p^{2} ) , \tau ( p^{3} ) , \tau ( p^{4} ) , \ldots$ for some prime numbers $p$.
Let's study the sequence $\tau ( p ) , \tau ( p^{2} ) , \tau ( p^{3} ) , \tau ( p^{4} ) , \ldots$ for some prime number $p$.
Suppose the possibility $\pm 1$ appears in the sequence $\tau ( p ) , \tau ( p^{2} ) , \tau ( p^{3} ) , \tau ( p^{4} ) , \ldots$ for some prime number $p$. There are 2-cases:
if $p$ divides $\tau ( p )$, then we easily proof that $p^{m} | \tau ( p^{m} )$ for all $m \geqslant 2$, the reason why is we can use induction and the property $\displaystyle{\tau ( p^{m+1} ) = \tau ( p ) \tau ( p^{m} ) - p^{11} \tau ( p^{m-1} ) }$ for all $m \geqslant 1$ and for every prime number $p$. So that implies $| \tau ( p^{m} ) | \neq 1$ for all $m \geq 2$, contradiction.
if $p$ does not divide $\tau ( p )$, then there are no defective Lucas numbers in the sequence $\tau ( p ) , \tau ( p^{2} ) , \tau ( p^{3} ) , \ldots = \pm 1$, i.e. there are no defective terms with $\tau(p^{m}) = \pm 1$.(Variations of Lehmer's conjecture for Ramanujan’s tau-function, look Lemma 2.1)
We defined that:
a prime number $l| u_{m}(\alpha_{p}, \beta_{p})$ is a primitive divisor of $\displaystyle{\tau ( p^{m-1} ) = u_{m}(\alpha_{p},\beta_{p}) = \frac{\alpha_{p}^{m} - \beta_{p}^{m}}{\alpha_{p} - \beta_{p}}}$ if $l \nmid (\alpha_{p} - \beta_{p})^2 u_{1}(\alpha_{p} , \beta_{p}) u_{2}(\alpha_{p} , \beta_{p}) \ldots u_{n-1}(\alpha_{p} , \beta_{p})$.
Those $u_{n}(\alpha_{p},\beta_{p})$, where $n>2$, without primitive divisor are called defective, where $\alpha_{p}, \beta_{p}$ are the roots of $x^{2} - \tau ( p ) x + p^{11} = ( x - \alpha_{p} ) ( x - \beta_{p} )$.
I cannot understand how can we lead from that: "there are no defective terms with $\tau(p^{m}) = \pm 1$" to a contradiction? Why that implies: "all of the terms $\tau(p) , \tau(p^{2}) , \tau(p^{3}), \ldots$ have a prime divisor"?
What is the motivation, behind that theorem $| \tau ( n ) | \neq 1$ for all $n \geq 2$ ?