Prove that $ n < 2^{n}$ for all natural numbers $n$.

Prove that
$
n < 2^{n}
$
for all natural numbers $n$.

I tried this with induction:
Inequality clearly holds when $n=1$.
Supposing that when $n=k$, $k<2^{k}$.
Considering $k+1 <2^{k}+1$, but where do I go from here?

Any other methods maybe?

Solutions Collecting From Web of "Prove that $ n < 2^{n}$ for all natural numbers $n$."

Proof by induction.

Let $n \in \mathbb{N}$.

Step $1.$: Let $n=1$ $\Rightarrow$ $n\lt2^{n}$ holds, since $ 1\lt 2$.

Step $2.$: Assume $ n \lt 2^{n}$ holds where $n=k$ and $k \geq 1$.

Step $3.$: Prove $n \lt 2^{n}$ holds for $n = k+ 1$ and $ k\geq 1$ to complete the proof.

$k \lt 2^{k}$, using step $2$.

$2\times k \lt 2\times2^{k}$

$ 2k \lt 2^{k+1}\quad(1)$

On the other hand, $k \gt 1 \Rightarrow k + 1 \lt k+k = 2k$. Hence $k+1\lt2k\quad(2)$

By merging results (1) and (2).

$k + 1 \lt 2k \lt 2^{k+1}$

$k + 1 \lt 2^{k+1}$

Hence, $ n \lt 2^{n}$ holds for all $ n \in \mathbb{N}$

Counting argument:

Let $S$ be a set with $n$ elements. There are $2^n$ subsets of $S$. There are $n$ singleton subsets of $S$. There is at least one non-singleton subset of $S$, the empty subset.

Since no-one’s posted it yet:

This is of course a special case of Cantor’s theorem: for any cardinal number $n$, $n<2^n$, and so in particular it’s true for all finite cardinals (aka naturals).

Note that $\displaystyle 2^n=(1+1)^n=1+\sum_{k=1}^{n}\binom{n}{k}>\binom{n}{1}=n$ holds for all $n\in \mathbb{N}$.

You can also prove this using the derivative. Since $n<2^n$ for $n=1$, and moreover:
$$1 < \log 2 \cdot 2^n$$
For all $n>1\in \mathbb{R}$, $n < 2^n$ for the same.

Hint:$$\large{1+z+z^2+\ldots+z^n=\dfrac{z^{n+1}-1}{z-1}}$$

$$2^n=(2^n-1)+1=(1+2+2^1+\ldots+2^{n-1})(2-1)+1\gt\underbrace{(1+1+\ldots+1)}_{n-1}+1=n$$

To add yet another answer, let us use AM/GM inequality. For $n\geq1$ one has
$$\frac{2^{n}-1}{n}=\frac{2^0+2^1+\ldots +2^{n-1}}{n}\geq \left(2^{0+1+\ldots+(n-1)}\right)^{\frac1n}=2^{\frac{n-1}{2}}\geq1,$$
and therefore $2^n-n\geq 1$.

If we assume $2^k>k$

$2^{k+1}=2\cdot 2^k>2\cdot k$ which we need to be $\ge k+1\iff k\ge 1$

We need to prove the claim true for $n=k+1$, where $k\ge1$. That is, we need to prove that $k+1<2^{k+1}$. Observe that:
$$ \begin{align*}
k+1&<2^k+1 & \text{by the induction hypothesis} \\
&<2^k+2 \\
&=2^k+2^1 \\
&\le2^k+2^k & \text{since } 1\le k \\
&= 2(2^k) \\
&= 2^{k+1}
\end{align*} $$
as desired.

Here’s your proof:

graphical comparison

… just kidding of course … kind of.

Well, you can actually easily show that the derivative $\frac{d}{dx}2^x=2^x \log(2)$ is greater than 1 for all $x\geq1$ (the break-even point is somewhere around 0.528766) and since 1 is the derivative of $f(x)=x$ of course, we just need to show that $2^x>x$ for $x=1$, i.e. that $2^1>1$ and we can deduce that this will always be the case because the gradient is always greater for $2^x$ than for $x$. And since it is true for all real numbers $\geq 1$ it’s of course also true for the natural numbers. Uncountably infinite overkill if you will but still an easy proof.

You can also go on to prove that $2^x>x$ for all real numbers. For $x$ smaller than the above-mentioned break-even point of $x=-\frac{\log(\log(2))}{\log(2)}\approx 0.528766$ the above argument is true just in reverse. The gradient of $x$ will always be greater than that of $2^x$. For $x=-\frac{\log(\log(2))}{\log(2)}$ itself it’s a matter of a simple calculation to show that $2^x>x$ since $2^{-\frac{\log(\log(2))}{\log(2)}}=\frac{1}{\log(2)}\approx 1.442695$.

Again, Wolfram Alpha has a nice visualization for this.

inequality plot

So tl;dr of this: at no point are the two functions closer than for $x=-\frac{\log(\log(2))}{\log(2)}\approx 0.528766$ (this means especially that they do not cross) and even there $2^x>x$.

Well $2^k + 1 < 2^{k+1}$ for $k \geq 1$ so $k + 1 < 2^k + 1 < 2^{k+1}$.

Assume there is $n$ kind of fruit and you can choose one of each; so you have $2^n$ options,
if you can only choose one fruit of all, you will have $n$ option,

In which scenario you have more option?!