Question: Show that $2^n>\frac{(n-1)n}{2}$ without using induction.

MY attempt :

$$1+2+3+...+(n-2)+(n-1)=\frac{n(n-1)}{2}$$

Since $2^n>n$,

$$2^n+2^n+2^n+...+2^n+2^n>\frac{n(n-1)}{2}$$

$$(n-1)2^n>\frac{n(n-1)}{2}$$

SO I'm getting an obvious fact :( 

How to do it without induction?

Answer: 
Suppose you have a finite set of $n$ elements. $2^n$ is the number of all possible subsets. $n(n-1)/2$ is the number of subsets of size exactly two, which is strictly less than the number of all possible subsets.
If you are allowed to use "well known facts"

WKF1: $2^n - 1=\sum\_{k=0}^{n-1} 2^k$.

WKF2: $\frac {n(n-1)}2 = \sum\_{k=0}^{n-1} k$

WKF3: $2^k > k$

WKF4: $b\_k > a\_k$ then $\sum\_{k=0}^n b\_k > \sum\_{k=0}^n a\_k$.

So ... $2^n> 2^n-1 =\sum\_{k=0}^{n-1} 2^k > \sum\_{k=0}^{n-1} k = \frac {n(n-1)}2$.

....

But.... I'm not sure I'd call that a "proof without induction" as the WKFs were taken for granted without proofs and, most likely, to prove them one would have used induction.