Prove $3 \cdot 5 \cdot 7 \cdot 11 \cdot prime_n = 2k + 1$

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It is known that any prime greater than 2 is odd.

How do I show the combinations of all primes greater than 2 is also odd, $2k+1$?

I tried using induction, but what is appropriate for $prime_n$?

$3 \cdot 5 \cdot 7 \cdot 11 \cdot 2k + 1 = 2k + 1$?

Can we use $2k + 1$ to replace $prime_n$ since we know it will be odd?