Prove $\sum_{k=0}^{n} \frac{x^k}{k!} < e^x, \forall x > 0$. 
Claim:
  $$\sum_{k=0}^{n} \frac{x^k}{k!} < e^x, \qquad \forall x > 0$$

Proof: 
Let $f(x)=e^x$. Then the Taylor series gives
$$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$
$$= \sum_{k=0}^{n} \frac{x^k}{k!} + \sum_{k=n+1}^{\infty} \frac{x^k}{k!}$$
Now, because $x>0$, $$\sum_{n+1}^{\infty} \frac{x^k}{k!} > 0$$
$$\implies f(x) > f(x) - \sum_{n+1}^{\infty} \frac{x^k}{k!}$$
$$\implies \sum_{k=0}^{n} \frac{x^k}{k!} = f(x) - \sum_{n+1}^{\infty} \frac{x^k}{k!} < f(x)$$
QED.
I would appreciate constructive criticism on this proof.  
 A: You are assuming $e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!}$. In this case you can  observe that the sequence 
$$f_n(x) := \sum_{k=0}^n \frac{x^k}{k!}, \quad x > 0$$
is strictly increasing, as $f_{n+1}(x)$ is $f_n(x)$ plus a positive number (as $x$ is positive). This implies that $(f_n(x))_{n=0}^{\infty}$ admits a limit, which is $e^x$. Using the definition of limit and that $(f_n(x))_{n=0}^{\infty}$ is an increasing sequence, if you fix $\varepsilon > 0$ then you can find $N \in \mathbb{N}$ such that 
$$0 < e^x-f_n(x) < \varepsilon \text{ if } n>N$$
and this implies in particular that
$$f_n(x) < e^x.$$
Suppose you don't know $e^x = \sum_k \frac{x^k}{k!}.$ Consider 
$$g_n(x):=e^x - \sum_{k=0}^n \frac{x^k}{k!}, \quad x \geq 0.$$
Observe that $g_n(0) = 0$ for any $n \in \mathbb{N}$. Assume by contradiction that there is an $N \in \mathbb{N}$ such that 
$$e^x \leq \sum_{k=0}^N \frac{x^k}{k!}$$
and further, that this is the least natural number for which this happens, that is 
$$e^x > \sum_{k=0}^{N-1} \frac{x^k}{k!}.$$
This is equivalent to saying that
\begin{equation}
g_N(x) \leq 0
\end{equation}
and
$$g_{N-1}(x) > 0.$$
But one can compute directly
$$g_N'(x) = g_{N-1}(x) > 0.$$
This fact and $g_n(0) = 0$ for every $n \in \mathbb{N}$ yield $g_N(x) > 0$ if $x > 0$, contradiction with $g_N(x) \leq 0$.
A: You argument is fine, but there also is a shortcut. By exploiting Taylor's theorem with integral remainder we have:
$$ e^x = \sum_{k=0}^{n}\frac{x^k}{k!}+\frac{1}{n!}\int_{0}^{x} e^t (x-t)^n\,dt\tag{A} $$
and it is pretty obvious that the last integral is positive if $x> 0$, since it is the integral of a continuous and positive function. 
