Inequality for the Binomial Theorem 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?

The Binomial Theorem says that if $k$ is a positive integer, then
$$(1+x)^k=1+\binom{k}{1}x+\binom{k}{2}x^2+\cdots +\binom{k}{k}x^k.$$
Note that if $x$ is positive, then each term in the binomial expansion above is positive.
So for $k\ge 2$, we have 
$$(1+x)^k\ge 1+kx+\binom{k}{2}x^2.$$
In particular, if we put $x=1/k$, we find that 
$$\left(1+\frac{1}{k}\right)^k\gt 1+k\cdot \frac{1}{k}=2.$$
Remark: The fact that $(1+x)^n \ge 1+nx$ for $x$ positive is called Bernoulli's Inequality. (It can be stretched to $x\ge -1$, but that is irrelevant for us.)
The Bernoulli Inequality can also be proved by a simple induction argument, we do not really need the Binomial Theorem.  For if we know that $(1+x)^n \ge 1+nx$, then $(1+x)^{n+1}=(1+x)^n(1+x)\ge (1+nx)(1+x)=1+(n+1)x+nx^2\ge 1+(n+1)x$.
