A nice easy problem

We know $x! \approx \sqrt{2\pi x}\times \left(\frac{x}{e}\right)^x$

Therefore, $\sqrt[x]{x!} \approx (2\pi x)^{\frac{1}{2x}} \times \frac{x}{e}$

As $x$ gets arbitrarily large, $\frac{1}{2x}$ goes to 0. If we plug in 2017 and 2018: $\frac{2018}{e}-\frac{2017}{e}=\frac{1}{e}$

Therefore, $\boxed{\sqrt[2017]{2017!}<\sqrt[2018]{2018!}}$ by approximately $\frac{1}{e}$ .

Note that $f(x) = \sqrt[2018]{2017!\cdot x}$ is a strictly increasing function of $x\geq 0$ . Also, if we set $G = \sqrt[2017]{2017!},$ then $G^{2017} = 2017!$ and $f(G) = \sqrt[2018]{G^{2017} \cdot G} = \sqrt[2018]{G^{2018}} = G,$ so it follows that $f(x) \left\{\begin{array}{c} < \\ = \\ >\end{array}\right\} G \iff x \left\{\begin{array}{c} < \\ = \\ >\end{array}\right\} G$ or, by setting $x=2018,$ $\sqrt[2018]{2018!} \left\{\begin{array}{c} < \\ = \\ >\end{array}\right\} \sqrt[2017]{2017!} \iff 2018 \left\{\begin{array}{c} < \\ = \\ >\end{array}\right\} \sqrt[2017]{2017!}$

To finish, note that $2017!$ is the product of $2017$ non-negative numbers all less than $2018,$ so $2018 = \sqrt[2017]{2018^{2017}} > \sqrt[2017]{2017!} \implies \boxed{\sqrt[2018]{2018!} > \sqrt[2017]{2017!}}$

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Remark: The first part of my solution actually yields a method to give more inequalities. For instance, using AM-GM, $\sqrt[2017]{2017!} < \frac{1+2017}{2} = \frac{2018}{2} \implies \sqrt[2018]{\frac{2018!}{2}} > \sqrt[2017]{2017!}$

From the pattern- 1<root2!,root2!<cuberoot3!,cuberoot3!<fourth root of4!

Suppose that $\sqrt[2018]{2018!}=\sqrt[2017]{2017!}=x$ . Then $x^{2018}=2018!$ and $x^{2017}=2017!$ , which means that:

$\frac{x^{2018}}{x^{2017}}=\frac{2018!}{2017!}\Leftrightarrow x=2018$

But if $x=2018$ then $x^{2018}=2018!$ and $x^{2017}=2017!$ are both obviously FALSE ( $2018^{2018}\neq 2018!\neq 2017!$ ), meaning that $\sqrt[2018]{2018!}\neq\sqrt[2017]{2017!}$ .

Next, let's assume that: $\sqrt[2018]{2018!}<\sqrt[2017]{2017!}\Leftrightarrow\frac{\sqrt[2018]{2018!}}{\sqrt[2017]{2017!}}<1$

And assign $x$ and $y$ in the following way: $y=\sqrt[2018]{2018!}\Leftrightarrow y^{2018}=2018! \\ x=\sqrt[2017]{2017!}\Leftrightarrow x^{2017}=2017!$ Then, our assumption means that: $\frac{y}{x}<1~\rightarrow ~\left(\frac{y}{x}\right)^{2018}<1\Leftrightarrow \\ \frac{y^{2018}}{x\cdot x^{2017}}<1\Leftrightarrow \\ x>\frac{2018!}{2017!}\Leftrightarrow x>2018$ But if $x>2018$ then $x^{2017}=2017!$ is (again) obviously FALSE (eg. $2019^{2017}\neq2017!$ ), meaning that the assumption $\sqrt[2018]{2018!}<\sqrt[2017]{2017!}$ it's also FALSE.

Therefore $\boxed{\sqrt[2018]{2018!}>\sqrt[2017]{2017!}~}$ .