Which-is-greater 1000^1001 or 1001^1000 ? Let’s simplify these number, to have a comparison criteria.

Comparing $1000^{1001}$ or $1001^{1000}$

Since, $\ln x$ is strictly increasing, we can also compare

\[\begin{aligned} \ln(1000^{1001})&= \ln (\exp (1001 \cdot\ln 1000))=1001 \cdot\ln 1000 \\ \ln(1001^{1000})&= \ln (\exp (1000 \cdot\ln 1001))=1000 \cdot\ln 1001 \end{aligned}\]

Divided by $\color{blue}{1000\times1001}$, we can compare

\[\begin{aligned} (1001 \cdot\ln 1000)/(\color{blue}{1000\times1001})&=\ln 1000/1000=f(1000) \\ (1000 \cdot\ln 1001)/(\color{blue}{1000\times1001})&=\ln 1001/1001=f(1001) \end{aligned}\]

Let’s study $f(x)=\ln (x)/x$

\[\forall x \in \mathbb{R}_{+}^{*},f^{\prime}(x)=\frac{1-\log (x)}{x^2}; \quad 0 = f^{\prime}(x) \Longleftrightarrow 0 =\frac{1-\log (x)}{x^2} \Longleftrightarrow x=e \text{ and } x\neq 0 \\\] \[\forall x \in \mathbb{R}_{+}^{*}, \quad 0 \lt f^{\prime}(x) \Longleftrightarrow 0 \lt \frac{1-\log (x)}{x^2} \Longleftrightarrow 0 \lt 1-\log (x)\Longleftrightarrow \log (x)\lt 1 \Longleftrightarrow x\lt e \\\] \[\forall x \in \mathbb{R}_{+}^{*}, \quad f^{\prime}(x) \lt 0 \Longleftrightarrow \frac{1-\log (x)}{x^2} \lt 0\Longleftrightarrow 1-\log (x)\lt 0 \Longleftrightarrow 1 \lt\log (x) \Longleftrightarrow e\lt x\] \[\color{green}{\forall x\in \left]e, +\infty \right[, f^{\prime}(x) \lt 0 \implies f \text{ is strictly decreasing}\implies \forall a, b \in \left]e, +\infty \right[, \quad a \lt b \implies f(a) \gt f(b)}\]

Taking $a=1000$ and $b=1001$

\[\begin{aligned} f(1000) &\gt f(1001) &\\ \ln (1000)/1000 &\gt \ln (1001)/1001 &\\ 1001 \cdot\ln 1000 &\gt 1000 \cdot\ln (1001) & \text{multiplied by }1000\times 1001\\ \exp (1001 \cdot\ln 1000) &\gt \exp (1000 \cdot\ln 1001)& \text{since } \exp \text{increasing} \end{aligned}\] \[\color{purple}{\boxed{1000^{1001} \gt 1001^{1000}}}\]