What is the smallest real number $m$ such that $n < m^n$ for all $n \geq 1$? I have taken a short look at this problem and found it to be much harder than expected to solve. Per the title, I am looking to find the smallest number $m\in\mathbb{R}$ such that the inequality $$n < m^n$$ is true for all $n\in\mathbb{R}$, $n \geq 1$. Using Python I was able to determine that $m\in [1.4446678610097, 1.4446678610098]$, suggesting the solution $m = e^{1/e}\approx 1.44466786100976\dots$. I assume this to in fact be the solution, and there there is a proof that I cannot come up with. Further research points to the Lambert W function, but the contents are beyond me at this point.
A solution/proof or explanation of this problem is appreciated.
 A: As was pointed out in the comments, there is no smallest such $m$.  To see why $m = e^{1/e}$ doesn't work, notice that with $n=e$ we get $m^n = (e^{1/e})^e = e = n$.
However, if you're willing to ask instead about the smallest positive $m$ for which $n \leq m^n$ for all $n \geq 1$, then $m = e^{1/e}$ is indeed correct.  First, observe that with $m$ positive the inequality $n \leq m^n$ is equivalent to $\ln(n) \leq n\ln(m)$, or equivalently $\frac{\ln(n)}{n} \leq \ln(m)$.
You can now use calculus to prove that the absolute maximum of the function $f(x) = \frac{\ln(x)}{x}$ on $(0, \infty)$ occurs at $x=e$, and at that value $f(e) = \frac{1}{e}$.  Thus we should choose $m$ so that $\ln(m) = \frac{1}{e}$, which gives us $m = e^{1/e}$, as you expected.
A: Essentially, you're looking for a proof that the maximum value of $x^{\frac1x}$ for $x>1$ occurs at $x=e$.
Let $y=x^{\frac1x}$. Then $x\ln(y)=\ln(x)$. Taking the derivative of both sides with respect to $x$, we get $$\ln(y)+x\cdot\frac1y\cdot \frac{dy}{dx}=\frac1x,$$ which rearranges to $$\frac{dy}{dx}=x^{\frac1x}\cdot\left(\frac1{x^2}-\frac{\ln(x)}{x^2}\right).$$ The derivative $\frac{dy}{dx}$ is equal to 0 for $x>1$ only at $x=e$, and we can confirm that $\frac{dy}{dx}$ indeed changes from positive to negative at $x=e$, thus proving that $y=x^{\frac1x}$ achieves its maximum value at $x=e$.
Then we have $$x^{\frac1x}\leq e^{\frac1e}\implies x\leq \left(e^{\frac1e}\right)^x$$ for all $x>1$.
