I want to show that $(1+ \frac{1}{k})^k \geq 2$ for say $k \geq 2$. Here is what I have so far:
By the Binomial Theorem we know $(1 + x)^k$ for $k \geq 2$ gives us:
$1^k + {k\choose 1}1^{k-1}x^1 + {k\choose 2}1^{k-2}x^2$ which yields:
$1 + kx + \frac{k(k-1)}{2}x^2$
We know substitue $x = \frac{1}{k}$:
$1 + 1 + \frac{1}{2} - \frac{1}{k}$
Back to the inequality we care about:
$\frac{5}{2} - \frac{1}{k} \geq 2$
Subtracting $2$ from both sides:
$\frac{1}{2} - \frac{1}{k} \geq 0$
Thus, this is true for all $k \geq 2$
Is this correct?