Prove $0.9999^{\!101}<0.99<0.9999^{\!100}$

$$0.99<0.9999^{100}$$ it's $$1-\frac{1}{100}<\left(1-\frac{1}{100^2}\right)^{100},$$ which is true by Bernoulli: $$\left(1-\frac{1}{100^2}\right)^{100}>1-100\cdot\frac{1}{100^2}=1-\frac{1}{100}.$$ $$0.9999^{101}<0.99$$ it's $$\left(1-\frac{1}{100^2}\right)^{101}<1-\frac{1}{100}$$ or $$\left(1-\frac{1}{100^2}\right)^{100}\left(1+\frac{1}{100}\right)<1$$ or $$\left(1-\frac{1}{100^2}\right)^{100}<1-\frac{1}{101},$$ which is true because $$\left(1-\frac{1}{100^2}\right)^{100}<1-\binom{100}{1}\cdot\frac{1}{100^2}+\binom{100}{2}\cdot\frac{1}{100^4}<1-\frac{1}{101}.$$ I used the following lemma.

Let $0<x<1$ and $n>2.$ Prove that: $$(1-x)^n<1-nx+\frac{n(n-1)}{2}x^2.$$

Proof.

We need to prove that $f(x)>0,$ where $$f(x)=1-nx+\frac{n(n-1)}{2}x^2-(1-x)^n.$$ Indeed, $$f'(x)=-n+n(n-1)x+n(1-x)^{n-1}$$ and $$f''(x)=n(n-1)-n(n-1)(1-x)^{n-2}=n(n-1)\left(1-(1-x)^{n-2}\right)>0.$$ Thus, $$f'(x)>f'(0)=0,$$ $$f(x)>f(0)=0$$ and the lemma is proven.

Now, take $x=\frac{1}{100^2}$ and $n=100.$