Divisibility proof through prime-factorization. I have to show that the following expression:
$$ \frac {1}{2} (3^{2^{n}}-1) $$
can be divided by $n-1$ different and odd prime numbers for every positive $n$ (I assume that $n \in \mathbb{N^*}$ )
So far I tried to prove it with the Lemma that claims, that every number $n \in \mathbb{N} :n \ge 2 $ has at least one prime factor, which seemed not to be utmost successful.
I am looking for a direction and for a better understanding of such proofs.
Thanks in advance for your help :)
 A: The fact in Tito Eliatron's question comment can be used in solving this problem. As stated there,
$$3^{2^{n}} - 1 = \left(3^{2^{n-1}} - 1\right)\left(3^{2^{n-1}} + 1\right) \tag{1}\label{eq1A}$$
You can repeat this factoring with $3^{2^{n-1}} - 1$ and each additional smaller power of $2$ factor down to $2^0 = 1$ to end up with
$$\begin{equation}\begin{aligned}
\frac{1}{2}(3^{2^{n}} - 1) & = \frac{1}{2}\left(3^{2^{0}} - 1\right)\left(3^{2^{0}} + 1\right)\left(3^{2^{1}} + 1\right)\ldots \left(3^{2^{n-1}} + 1\right) \\
& = 4\prod_{i=1}^{n-1}\left(3^{2^{i}} + 1\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Each factor in the product above has $3$ with an exponent $\ge 2$. Consider any one of them to be just a power of $2$, say for some exponent $2^i = j$. Thus, for some $k \ge 2$, we would have
$$3^{j} + 1 = 2^k \implies 2^k - 3^{j} = 1 \tag{3}\label{eq3A}$$
However, Catalan's conjecture, which was proven by Preda Mihăilescu in $2002$, states the only solution for a difference  of positive integers to a power of at least $2$ to be $1$ are for $3^2 - 2^3 = 1$. Thus, this means \eqref{eq3A} will never occur.
As such, each $3^{2^{i}} + 1$ for $i \ge 1$ must have at least one odd prime factor. Also, all of the odd prime factors in $3^{2^{i}} + 1$ for any $i$ are distinct from those of any other value of $i$. To see this, consider for any $i \ge 1$ an odd prime $p \mid 3^{2^{i}} + 1$. This gives, for any integer $m \gt 0$,
$$\begin{equation}\begin{aligned}
3^{2^{i}} & \equiv -1 \pmod{p} \\
3^{2^{i}\left(2^{m}\right)} & \equiv \left(-1\right)^{2^m} \pmod{p} \\
3^{2^{i+m}} & \equiv 1 \pmod{p}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
However, $p \mid 3^{2^{i+m}} + 1$ requires $3^{2^{i+m}} \equiv -1 \pmod{p}$ which means, from \eqref{eq4A}, that $-1 \equiv 1 \pmod{p}$, but this is never true for any odd prime $p$.
This confirms there's at least one distinct odd prime factor among each of the $n - 1$ factors of $3^{2^{i}} + 1$ in \eqref{eq2A}.
