Prove that $e$ exists and has a value of $2.71828$ We've been working on proving different mathematical formulas, statements and constants such as the existence of the $sin$ function. Now we need to prove that $\exp$ exists and has the value of $2.71828...$. Here we use $\exp$ and not $e$, as the solution to the unique problem of $f'(x)=f(x), f(0)=1$. 
I started of by stating that $$\exp(x)=\sum_{k=0}^n \frac{x^k}{k!} + R_n (x)$$
I got stuck right after that, but I assume first I state that $x∈[a,b]$. Then I need to work out the value of $R_n(x)$ as $$|R_n(x)|=|\exp(ζ)\frac{x^{n+1}}{(n+1)!}|≤\exp(b)\frac{|x|^{n+1}}{(n+1)!}\to \ 0$$
I need help continuing on from there. 
 A: The problem with using Taylor's theorem to bound the error term is that you need to already have an estimate for how large $\exp(x)$ gets between $0$ and $1$.
It is easier to estimate the error term directly by bounding the rest of the series by a geometric series:
$$ \sum_{k\ge n} \frac{1}{k!} \le \frac{1}{n!} \sum_{h\ge 0}\Bigl(\frac{1}{n+1}\Bigr)^h = 
\frac{1}{n!} \cdot \frac{1}{1-1/(n+1)} =
\frac1{n!} (1+1/n) $$
so just start summing the series, and you can stop as soon as the partial sum is at least $2.71828$ and the last term takes you at most $\frac{n}{n+1}$ of the way to $2.71829$.
About 10 terms should be enough, so doing the calculations to 7 decimal places will be sufficient to prevent rounding errors from upsetting the result.
A: To find the solution of $f'(x)=f(x)$, we want to solve the first-order linear ODE $$y'-y=0$$ where $y=f(x)$. Now $P(x)=-1$ and $Q(x)=0$ so the integrating factor is $$e^{\int P(x)\,dx}=e^{\int-1\,dx}=e^{-x}$$ Then $$ye^{-x}=\int e^{-x}Q(x)\,dx=C$$ where $C$ is a constant so $$\boxed{y=Ce^x}$$ is the only solution.
