Why is $\prod\limits_{p\le Y}\left(1+\frac{1+2e}{p}\right)\le(\log Y)^{10}$

Question

Why is $\prod\limits_{p\le Y}\left(1+\frac{1+2e}{p}\right)\le(\log Y)^{10}$

Can I bring $(1+2e)$ to the exponent and write $\prod\limits_{p\le Y}\left(1+\frac{1}{p}\right)^{1+2e}$ which is lesser than $7$, so is $\prod\limits_{p\le Y}\left(1+\frac{1}{p}\right)\le (\log Y)$ ?

Well In fact I should replace ''$\le$'' by $\ll$ but I think it is still true.

Edit: $p$'s are primes, Yes

Answer 1

Taking the log: $$ \log \left( \prod_{p \leq Y} (1+ \frac{1+2e}{p}) \right) = \sum_{p \leq Y} \log( 1+ \frac{1+2e}{p}) \overset{(*)}{\leq} \sum_{p \leq Y} \frac{1+2e}{p} = (1+2e) \sum_{p \leq Y} \frac{1}{p} $$ $(*)$ Since $\log(1+x) \leq x, \ \forall x$

And the last sum is known to be asymptotic to $\sim \log \log Y$.

Since $1+2e < 10$ we have the claim.