How to show that $2^n > n$ without induction

Question

I'm solving exercises about Pascal's triangle and Binomial theorem, and this problem showed up, however I don't have any clue on how to solve it

The sum of ${n\choose p}$ from $p=0$ to $n$ is the same thing as $(1+1)^n=2^n$, how can I use this information? Maybe comparing with another summation that equals to n?

Answer 1

Use Bernoulii inequality, which is true for all $x>-1$: $$ (1+x)^n\geq 1+nx$$ so $$(1+1)^n \geq 1+n\cdot 1 >n$$

Answer 2

Maybe comparing with another summation that equals to $n$?

For any $p=0,1,\dots,n$, there is at least one way to choose $p$ things from a list of $n$ things. Thus $\binom{n}{p} \ge 1$, so

$$ 2^n = \sum_{p=0}^n \binom{n}{p} ≥ \sum_{p=0}^n 1 = n+1 > n.$$

Answer 3

Note that $2^n$ is the number of subsets of $[n]=\{1,\dotsc,n\}$. There are $n$ subsets of $[n]$ with size $1$. There is at least one subset of $[n]$ which is not a singleton (namely the empty set). Hence $$ 2^n>n $$ for $n\geq 1$.

Answer 4

hint

Consider $x\mapsto \frac{\ln(x)}{x}$ for $x\ge 1$.

$$f'(x)=\frac{1-\ln(x)}{x^2}$$

the maximum if $f(e)=\frac{1}{2}<\ln(2)$.

thus

$$\ln(x)<x\ln(2)$$

Answer 5

The Binomial Theorem says $$ \begin{align} 2^n &=(1+1)^n\\ &=\binom{n}{0}1^0+\binom{n}{1}1^1+\dots\\ &\ge1+n\\[9pt] &\gt n \end{align} $$