what is $2\cdot 4\cdot 6\cdot 8 \cdot \ldots \cdot (2k+2)$?

Question

I know that $2\cdot 4\cdot 6\cdot 8 \cdot \ldots \cdot (2k)$ is $2^kk!$ but what is the value of these terms up to the $(2k+2)^\text{th}$ term?

Answer 1

$$2\cdot 4\cdot 6\cdot 8 \cdot \ldots \cdot (2k+2)~=~(2\cdot\underline1)(2\cdot\underline2)(2\cdot\underline3)(2\cdot\underline4)\ldots\big(2\cdot[k+1]\big)~=~2^{k+1}(k+1)!$$