e is irrational Prove that e is an irrational number.
Recall that $\,\mathrm{e}=\displaystyle\sum_{n=0}^\infty\frac{1}{n!},\,\,$ and assume $\,\mathrm{e}\,$ is rational, then
$$\sum\limits_{k=0}^\infty \frac{1}{k!} = \frac{a}{b},\quad \text{for some positive integers}\,\,\, a,b.$$
so
$$b\sum\limits_{k=0}^\infty \frac{1}{k!} =a$$
or
$$
b\left(1+1+\frac{1}{2} + \frac{1}{6} +\cdots \right)= a.
$$
Where can I go from here?
 A: Hints. 
We first show that $2<\mathrm{e}<3$ (see below), and hence $\mathrm{e}$ is not an integer.
Next, following up OP's thought, assuming $\mathrm{e}=a/b$, we multiply by $b!$ and we obtain
$$
\sum_{k=0}^\infty \frac{b!}{k!}=a\cdot (b-1)! \tag{1}
$$
The right hand side of $(1)$ is an integer.
The left hand side of $(1)$ is of the form
$$
\sum_{k=0}^b \frac{b!}{k!}+\sum_{k=b+1}^\infty \frac{b!}{k!}= p+r.
$$
Note that $p=\sum_{k=0}^b \frac{b!}{k!}$ is an integer, while $$
0<r=\sum_{k=b+1}^\infty \frac{b!}{k!}=\frac{1}{b+1}+\frac{1}{(b+1)(b+2)}+\cdots<\sum_{k=1}^\infty \frac{1}{(b+1)^k}=\frac{1}{b}<1.
$$
Note. The fact that $\mathrm{e}\in (2,3)$ can be derived from the inequalities 
$$ 
\left(1+\frac{1}{n}\right)^{\!n}<\mathrm{e}<\left(1+\frac{1}{n}\right)^{\!n+1},
$$
for $n=1$ for the left inequality and $n=5$ for the right inequality.
A: While it may not lead to a better formal proof than other approaches, I find the following quite intuitive. It also compares nicely to Euler's original proof using the continued fraction representation of$~e$.
It is easy to see that every rational number $\alpha\in(0,1)_\Bbb Q$ can be uniquely expressed as
$$\alpha = \frac{c_1}{1!}+\cfrac{c_2}{2!}+\cdots+\cfrac{c_k}{k!}
 \quad\text{with $k>1$, integers $0\leq c_i<i$ for $i=0,\ldots,k$, and $c_k\neq0$.}
$$
One can interpret this as the expression of $\alpha$ in the
fractional counterpart of the factorial number system.
Writing this as $\alpha=(c_1,c_2,\ldots,c_k)_!$ ones has the familiar property of finite decimal representations (without integer part), that the order of two such numbers is given by lexicographic comparison of their representations $(c_1,c_2,\ldots,c_k)$.
Now $e-2$ is the limit of the rational numbers of the form $(0,1,\ldots,1)_!$ as the number of digits $1$ goes to infinity (the "$-2$" removes to two integral terms $\frac1{0!}$ and $\frac1{1!}$). Clearly this cannot converge to any number with a finite representation $(c_1,c_2,\ldots,c_k)_!$.

If one would want to turn this into a formal argument, one could argue as follows. Once the "digit" in position $k+1$ has become $1$, all further numbers in the sequence lie in the closed interval between the $k+1$ digit numbers $(0,1,\ldots,1,1)_!$ and $(0,1,\ldots,1,2)_!$, which interval does not contain any number with a $k$-digit representation $(c_1,c_2,\ldots,c_k)_!$, and in particular the sequence cannot converge to such a number.
A: I like the following mild variant, which is less popular. Equivalently, we prove that $e^{-1}$ is irrational. Suppose to the contrary that $e^{-1}=\frac{m}{n}$ where $m$ and $n$ are integers with $n\gt 0$.
We have 
$$e^{-1}=\sum_{k=0}^{n}\frac{(-1)^k}{k!}+\sum_{k=n+1}^\infty \frac{(-1)^k}{k!}.$$
Multiply through by $n!$. We get that
$$n!\sum_{k=n+1}^\infty \frac{(-1)^k}{k!}$$
must be an integer. 
However, by the reasoning that leads to the Alternating Series Test, we have
$$0\lt \left|n!\sum_{n+1}^\infty \frac{(-1)^k}{k!}\right|\lt \frac{n!}{(n+1)!}\lt 1.$$
The (small) advantage is that the estimation of the tail is easier than when we use the series expansion of $e$. 
A: Here is yet another one, which is one of my favorite irrationality/transcendence proofs :
The confluent hypergeometreic series 
$$_{0}F_{1}(k; z) = \sum_{n = 0}^\infty \frac1{(k)_n} \frac{z^n}{n!}$$
Satisfies the more-or-less easily verifiable identity 
$$_0F_1(k-1;z) - {}_0F_1(k; z) = \frac{z}{k(k-1)}{}_0F_1(k+1;z)$$
Iterating this, one ends up with the continued fraction 
$$\frac{{}_0F_1(k+1;z)}{k{}_0F_1(k;z)} = \frac1{k+\cfrac{z}{k+1+\cfrac{z}{k+2+\cfrac{z}{k+3+\cdots}}}}$$ 
Now note that $_0F_1(3/2;x^2/4) = \cosh(x)$ and $x \,{}_0F_1(3/2;x^2/4) = \sinh(x)$, hence applying the above one has the pretty well-known continued fraction
$$\tanh(x) = \cfrac{x/2}{\frac{1}{2} + \cfrac{\frac{x^2}{4}}{\frac{3}{2} + \cfrac{\frac{x^2}{4}}{\frac{5}{2} + \cfrac{\frac{x^2}{4}}{\frac{7}{2} + \cdots}}}} = \cfrac{x}{1 + \cfrac{x^2}{3 + \cfrac{x^2}{5 + \cfrac{x^2}{7 + \cdots}}}}$$
Note, however, that $$\tanh(x) = \frac{\exp(x) - \exp(-x)}{\exp(x) + \exp(-x)}$$ but since the continued fraction above is not finite, $\tanh(1)$ is not rational. Hence $e$ is not rational either.
