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$$
\left(1+\dfrac{1}{n}\right)^n < e < \left(1+\dfrac{1}{n}\right)^{n+1}
$$
We will assume that $n \in \mathbb{N}$, $n > 0$.
Left Side $(1+\dfrac{1}{n})^n < e$
We already know that$e = \text{lim}_{n \rightarrow \infty} (1 + \dfrac{1}{n})^n$. In that limit, $(1 + \dfrac{1}{n})^n$ is effectively the left side of the above inequality.
This implies that $(1+\dfrac{1}{n})^n$ has an asymptote at $e$, as $n\rightarrow\infty$.
The only way that the inequality will hold is if $(1+\dfrac{1}{n})^n$ increases monotonically. Put another way, for all $n$, there exists a $n + 1$ such that $(1+\dfrac{1}{n+1})^{n+1} > (1+\dfrac{1}{n})^{n}$. If that's the case, then we know that for any $n$, $(1+\dfrac{1}{n})^n < e$
We shall prove by induction that $(1 + \dfrac{1}{n})^n > (1 + \dfrac{1}{n - 1})^{n-1}$, can be algebraically manipulated to yield the Bernoulli's Inequality.
By Bernoulli's Inequality, let $x = -\dfrac{1}{n^2}$, then the left hand side is $(1 + x)^n$, and, the right-hand side is $1 + nx$, thus giving us the inequality
Same procedure, again. We want to re-interpret the above resulting inequality to Bernoulli's Inequality, of the form $(1 + x)^n > 1 + xn$.
We want to reinterpret $x$ as $\dfrac{1}{n(n+2)}$, and $n$ as $n + 2$. The left hand side is trivially equivalent. For the right hand side where $1 + xn$ is congruent with $1 + \dfrac{1}{n(n+2)}(n+2)$.
Or, in words, that $\left(1+\dfrac{1}{n}\right)^{n+1}$ is monotonically decreasing for all $n$, and $n+1$.
Now because $\left(1+\dfrac{1}{n}\right)^{n+1}$ has an asymptote at $e$ as $n \rightarrow \infty$, and that it is monotonically decreasing, then for all $n$, $\left(1+\dfrac{1}{n}\right)^{n+1} > e$.