Proving the Inequality of e^x Using Taylor's Theorem

[AaronEx]

Homework Statement

Show that if $0 \le x \le a$, and $n$ is a natural number, then $$1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!} \le e^x \le 1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!}+\frac{e^ax^{n+1}}{(n+1)!}$$

Homework Equations

I used Taylor's theorem to prove $e^x$ is equal to the LHS of the inequality plus an error term $\frac{e^cx^{n+1}}{(n+1)!}$, where the series is of order $n$ centered at $x=0$.

The Attempt at a Solution

With the expansion of $e^x$ in the relevant equations section, I proved that, since the error term is greater than or equal to zero, the inequality holds true. Is this correct? As for the RHS inequality, I represented $e^x$ as the expansion again, but this time stating that the expansion is on the interval $[0,a]$, and so I argued that the RHS was larger than $e^x$ since $e^c \le e^a$. Is this true as well? Have I left anything relevant out?

 

[mfb]

Looks fine.