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: 
\begin{eqnarray}
2^0 &\geq& 1 \\
2^1 &\geq& 2 \\
2^2 &>& 3 \\
&\vdots& \\
2^{n-1} &>& n-1 \quad \text{for } n>1 \\
\end{eqnarray}
If you add everything up, you get (for $n>1$):
$$
2^n -1 = 2^0+2^1+...+2^{n-1} > 1+2+3+...+(n-1) = \frac{n(n-1)}{2}
$$
which means that for $n>1$
$$
2^n > 1+\frac{n(n-1)}{2} > \frac{n(n-1)}{2}
$$
We can manually check that this inequality also holds for $n=1$
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.