Prime numbers $p$ such that $p-1$ has no small odd factors

Question

I am interested in a (simple) proof of this result:

There exists $\alpha>0$ and infinitely many prime numbers $p$ such that $p-1$ does not have any odd prime factor smaller than $p^\alpha$.

I know that one can do much better by using complicated sieves, such as the Brun-Hooley sieve or the "lower bound sieve" as seen in Cojocaru & Murty, chapter 10, for instance, but I am only interested in this weaker form and a simpler proof.

I tried using Brun's pure sieve and the Bombieri-Vinogradov theorem but it didn't really work out. If my original question is too complicated, I would also appreciate a simple proof of this (slightly weaker) result:

There exists $C$ and infinitely many prime numbers $p$ such that $\omega(p-1)<C$.

Edit: I wish to clarify that I intend to find a proof with a reduced degree of technicality when it comes to the sieving part. I do allow the usage of powerful theorems, such as Bombieri-Vinogradov.

Answer 1

This isn't quite a proof yet, but I'm pretty sure you can get $α>0.5$ because for most primes $q$ there exists a prime $p$ such that $p=kq+1$ for $k \leq q$. For any given $q$, we expect roughly $q/2\ln (q)$ of the $k$ values to yield a prime $p$.

I'm actually pretty sure this let's you get $α$ arbitrarily close to $1$.