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An early paragraph in Apostol's Calculus says that "with a little effort" we can show that $2 \leq \left(1 + \frac{1}{n}\right)^n$ for all integers $n \geq 1$. This is before induction or the binomial theorem are introduced, but after the field and order axioms. Is there a way to prove this without induction or the binomial theorem?

Straight inequality manipulation starts with $n \geq 1$, then $1/n \leq 1$, so $1 + 1/n \leq 2$. Both sides are positive, so raising both sides to the $n$th power gives $$\left(1 + \frac{1}{n}\right)^n \leq 2^n$$ Which is almost the opposite of what we want.

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Note that by making "explicitly" the computation you get $$ \left(1+\frac1n\right)^n=\underbrace{\left(1+\frac1n\right)\cdots\left(1+\frac1n\right)}_{n \ \rm times}=1+\underbrace{\frac1n+\cdots+\frac1n}_{n \ \rm times}+\cdots=2+\cdots>2 $$ for $n>1$.

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    $\begingroup$ Nice, though there is hidden a little induction there. +1 $\endgroup$ – DonAntonio Feb 28 '16 at 2:44
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If $x \geqslant 1$, then $x^n - 1 = (x - 1)(x^{n-1} + x^{n - 2} + \cdots + x + 1) \geqslant n(x - 1)$.

Taking $x = 1 + \frac{1}{n}$, we get $\left(1 + \frac{1}{n}\right)^n - 1 \geqslant 1$.

Arguably this involves a hidden appeal to mathematical induction.

(But is there anything in analysis that doesn't?)

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For $n = 1$ its true, Consider $n \geq 2$, $x = \dfrac{1}{n}$, we prove: $2^x \leq 1+x$ when $0 < x \leq \dfrac{1}{2}$. Consider $f(x) = 1+x-2^x$, then $f'(x) = 1-2^x\ln 2 > 1- \sqrt{2}\ln 2 = 0.019 > 0\Rightarrow f(x) > f(0) = 0.$Thus its proven.

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