How do I show that this function is always $> 0$ 
Show that $$f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +
 \frac{x^4}{4!} > 0 ~~~ \forall_x \in \mathbb{R}$$

I can show that the first 3 terms are $> 0$ for all $x$:
$(x+1)^2 + 1 > 0$
But, I'm having trouble with the last two terms. I tried to show that the following was true:
$\frac{x^3}{3!} \leq \frac{x^4}{4!}$
$4x^3 \leq x^4$
$4 \leq x$
which is not true for all $x$.
I tried taking the derivative and all that I could ascertain was that the the function became more and more increasing as $x \rightarrow \infty$ and became more and more decreasing as $x \rightarrow -\infty$, but I couldn't seem to prove that there were no roots to go with this property.

Hint:
$$f(x) = \frac{1}{4} + \frac{(x + 3/2)^2}{3} +\frac{x^2(x+2)^2}{24}$$
