Compare numbers with big powers Compare $2019^{2020}$ and $2020^{2019}$.
I know that $2019^{2020}$ is greater than $2020^{2019}$ but I couldn't prove it. I tried proving $(\frac{2019}{2020})^{2019}.2019>1$ but without success.  
 A: In general, for $n\geq 1$, you have
$$
n^{n+1} = n\cdot n^n
$$
while 
$$
(n+1)^n = \left(n\cdot\left(1+\frac{1}{n}\right)\right)^n
= \left(1+\frac{1}{n}\right)^n\cdot n^n
$$
The question then boils down to decide 

Which is bigger: $n$ or $\left(1+\frac{1}{n}\right)^n$ ?



*

*we expect it to be $n$, when $n$ is large, since $\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e \approx 2.7$

*and indeed, we can use the standard inequality $\ln(1+x) \leq x$ (for any $x> -1$) to say that 
$$
\left(1+\frac{1}{n}\right)^n 
= e^{n \ln \left(1+\frac{1}{n}\right)}
\leq e^{n \cdot \frac{1}{n}} = e^1 = e
$$
Therefore, since $e< 3$ we have
$$
\left(1+\frac{1}{n}\right)^n \leq n, \qquad \forall n \geq 3
$$
proving that 
$$
  \boxed{(n+1)^n < n^{n+1} \qquad \forall n \geq 3}
$$

What you want is for $n=2019$.
A: $$\frac {2020^{2019}}{2019^{2020}}=\frac {1}{2019} \left[1+\frac {1}{2019}\right]^{2019}=$$ 
$$=\frac {1}{2019}\sum_{j=0}^{2019}\binom {2019}{j}\frac {1}{2019^j}=$$
$$=\frac {1}{2019}\left[2+\sum_{j=2}^{2019}\binom {2019}{j}\frac {1}{2019^j}\right]<$$
$$<\frac {1}{2019}\left[2+\sum_{j=2}^{2019}\frac {1}{j!}\right]<$$
$$<\frac {1}{2019}\left[2+\sum_{j=2}^{2019}\frac {1}{2^{j-1}}\right]=$$
$$=\frac {1}{2019}[2+(1-2^{-2018})]<$$ $$<\frac {3}{2019}<1. $$ The transition from the 3rd line to the 4th line is because if $2\le j\le 2019$ then $$\binom {2019}{j}\frac {1}{2019^j}=\frac {1}{j!}\prod_{n=1}^j \frac {2019-(n-1)}{2019}=\frac {1}{j!}\prod_{n=1}^j\left(1-\frac {n-1}{2019}\right)<\frac {1}{j!}.$$
A: Since you already received good and simple answers, let me show another way.
You want to compare
$x^{x+1}$ to $(x+1)^x$ for large values of $x$, that is to say that you want to know if
$$A =\frac{x^{x+1}}{(x+1)^x}$$ is or not larger than $1$, that is to say if $\log(A)$ is or not positive.
Then
$$\log(A)=(x+1)\log(x)-x \log(x+1)=(x+1)\log(x)-x \left(\log \left(1+\frac{1}{x}\right)+\log (x)\right)$$that is to say
$$\log(A)=\log (x)-x \log \left(1+\frac{1}{x}\right)$$ Using the classical Taylor expansion of $\log(1+\epsilon)$, we then have
$$\log(A)=-1+\log(x)+\frac{1}{2 x}-\frac{1}{3x^2}+O\left(\frac{1}{x^3}\right)$$ Continuing with Taylor
$$A=e^{\log(A)}=\frac{x}{e}+\frac{1}{2 e}-\frac{5}{24 e x}+O\left(\frac{1}{x^2}\right)$$ 
So
$$A \sim \frac{2x+1}{2e}$$ Using $x=2019$, this would give $A\sim\frac{4039}{2 e}\approx 742.93253$ while the "exact" value would be $\approx 742.93249$
