If p are primes and m any integer, show that $\displaystyle \prod_{p\leq m+1} p \leq 4^m$, i was told it can be done by induction but still can't prove it...
I'll let you verify the base case.
So assume true for all $\displaystyle i\leq m $ and let's use this to show the case for $\displaystyle m+1 $ is true too.
Now, we have that $\displaystyle 2^{m+1}\geq{m+1 \choose \lfloor (m+1)/2 \rfloor} \geq \prod_{\lfloor (m+1)/2 \rfloor \leq p <m+1} p $
By our inductive hypothesis, $\displaystyle \prod_{p\leq\lfloor (m+1)/2\rfloor} p \leq 4^{\lfloor (m+1)/2\rfloor-1} = 2^{2\lfloor (m+1)/2\rfloor-2} $
So $\displaystyle 2^{m+1}\cdot2^{2\lfloor (m+1)/2\rfloor-2}\geq\prod_{p\leq m+1} p $
Notice that $\displaystyle m+1+2\lfloor (m+1)/2\rfloor-2 \leq m+1+2 (m+1)/2-2 = 2m $
Thus $\displaystyle 2^{m+1}\cdot2^{2\lfloor (m+1)/2\rfloor-2}\leq 2^{2m} = 4^m $.