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Yemon Choi
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You can also prove that $$\displaystyle\lim_n\sqrt[p_n]{\prod_1^n p_i} = e$$$$\displaystyle\lim_n\left(\prod_1^n p_i\right)^{1/p_n} = e$$

(where $p_i$ is the $i$-th prime number and $e$ is Euler's exponential number)

You can also prove that $$\displaystyle\lim_n\sqrt[p_n]{\prod_1^n p_i} = e$$

(where $p_i$ is the $i$-th prime number and $e$ is Euler's exponential number)

You can also prove that $$\displaystyle\lim_n\left(\prod_1^n p_i\right)^{1/p_n} = e$$

(where $p_i$ is the $i$-th prime number and $e$ is Euler's exponential number)

You can also prove that $\displaystyle \\ \lim_n \\ \\ \sqrt[p_n]{\prod_1^n p_i} = e$$$\displaystyle\lim_n\sqrt[p_n]{\prod_1^n p_i} = e$$

(where $p_i$ is the $i$-ethth prime number and $e$ is Euler's exponential number)

You can also prove that $\displaystyle \\ \lim_n \\ \\ \sqrt[p_n]{\prod_1^n p_i} = e$

(where $p_i$ is the $i$-eth prime number and $e$ is Euler's exponential number)

You can also prove that $$\displaystyle\lim_n\sqrt[p_n]{\prod_1^n p_i} = e$$

(where $p_i$ is the $i$-th prime number and $e$ is Euler's exponential number)

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Jose Brox
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You can also prove that $\displaystyle \\ \lim_n \\ \\ \sqrt[p_n]{\prod_1^n p_i} = e$

(where $p_i$ is the $i$-eth prime number and $e$ is Euler's exponential number)