Search for simpler (other) idea to show $(1.001)^{1000}>2$ The below question asked me to solve: (without using logarithm) which one is bigger $$2 \ , \ (1.001)^{1000}$$ I use two Idea to show $(1.001)^{1000}$ is bigger.

First: $$\lim_{n \to \infty}\left(1+\frac 1n\right)^n=e=2.71...\\$$and $1000$ is big enough to say $$(1.001)^{1000} \sim e>2$$
Second: Using binomial expansion $$(1.001)^{1000}=(1+0.001)^{1000}=\\1+1000\times 1^{999} \times (0.001) +\frac {1000\times 999}{2}1^{998}(.001)^2+...=\\1+1+\frac{999}{2000}+...>2$$   Now ,my question is there any simpler or other Idea to show $(1.001)^{1000}>2$
Thanks in advance for any hint.
 A: According to Bernoullli's inequality, we have $(1+x)^r\ge1+xr$. for this problem consider $x=0.001$ and $r=1000$.
Note: Here $(1+x)^r=1+xr$ is not the case because $r\neq0$
A: The inequality is equivalent to
$$2^{\frac 1{1000}} <1+\frac 1{1000} $$
But this is true according to GM-AM:
$$2^{\frac 1{1000}} = \sqrt[1000]{1^{999}\cdot 2}\stackrel{GM-AM}{<}\frac{999+2}{1000} = 1+\frac 1{1000}$$
A: By elementary arithmetic:
$$1.001^2=1.002001\ge1.002$$
$$1.001^4=1.004006004001\ge1.004$$
$$\cdots$$
$$1.001^{128}>1.128.$$
$$1.001^{256}>1.256.$$
$$1.001^{512}>1.512.$$
Finally,
$$1.001^{1000}>1.001^{512+256+128}>1.512\cdot1.256\cdot1.128>2.$$
A: Let's pretend we know nothing of mathematical induction, so we can't just say $(1+x)^n\ge1+nx$ for all $n\ge1$ (and $x\ge-1$), but let's assume we can do enough algebra to know that
$$(1+x)^2=1+2x+x^2\gt1+2x$$
and
$$(1+x)^5=1+5x+10x^2+10x^3+5x^4+x^5\gt1+5x$$
for all $x\gt0$.  Then we can show
$$(1+x)^{10}\gt(1+5x)^2\gt1+10x$$
from which it follows that
$$(1+x)^{100}\gt(1+10x)^{10}\gt1+100x$$
and, finally,
$$(1+x)^{1000}\gt(1+100x)^{10}\gt1+1000x$$
at which point we can plug in $x=1/1000$ to conclude
$$1.001^{1000}=\left(1+{1\over1000}\right)^{1000}\gt1+{1000\over1000}=2$$
A: You can us bernoulli inequality
$(1,001)^{1000}=(\frac{1001}{1000})^{1000}=$
$(1+\frac{1}{1000})^{1000}\geq  1+\frac{1} {1000}×1000$
A: $$f(x):=\left(1+\frac1x\right)^x$$ is a strictly growing function of $x$, and $f(1)=2$.

Indeed for $x>0$,
$$(\log(f(x))'=\log\left(1+\frac1x\right)-\frac1{x+1}>0$$
because
$$(\log(f(x))''=-\frac1{x(x+1)^2}<0$$ and the first derivative decreases to $0$.
