infinite root product

$\begin{aligned} L & = \lim_{n \to \infty} \prod_{i=0}^n \left(1+\frac in\right)^\frac 1n \\ & = \lim_{n \to \infty} \prod_{i=0}^n \exp \left(\ln \left(1+\frac in\right)^\frac 1n \right) & \small \blue{\exp (x) = e^x} \\ & = \exp \left(\blue{\lim_{n \to \infty} \frac 1n \sum_{i=0}^n \ln \left(1+\frac in\right)} \right) & \small \blue{\text{By Riemann sums}} \\ & = \exp \left(\blue{\int_0^1 \ln(1+x)\ dx} \right) \\ & = \exp \left((1+x)\ln(1+x) - x \bigg|_0^1 \right) \\ & = \exp \left(2\ln 2 - 1 \right) \\ & = \frac 4e \approx \boxed{1.412} \end{aligned}$

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Reference: Riemann sums

product series of the form $\displaystyle\lim_{n \to \infty} \displaystyle \prod_{i=0}^{n} \left( a +\dfrac{i\times b}{n} \right) ^{\dfrac{1}{n}}$

can be written as follows $\dfrac{\left( a+b\right) \left( \dfrac{b}{a} +1 \right)^{\dfrac{a}{b}}}{e}$

in the special case above a=1 and b=1

$\dfrac{\left( 2\right) \left( 2 \right)^{1}}{e}$ = $\dfrac{4}{e}$

edit: this form can be derived using riemann sums here