Prove that (1+ dfrac1x)^x < e < (1+ dfrac1x)^(x+1)

Prove that $\left(1+\dfrac1x\right)^{x} \lt e \lt \left(1+\dfrac1x\right)^{x+1}$ for every $x\gt 0$.

Step 1: Take logarithms

All three quantities are positive and the logarithm is increasing, so the double inequality is equivalent to

\[ x\ln\left(1+\frac1x\right) \lt 1 \lt \bigl(x+1\bigr)\ln\left(1+\frac1x\right). \tag{1} \]

Step 2: Substitute $t = 1/x$

With $t = \dfrac1x\gt 0$, and multiplying through by $t\gt 0$, (1) becomes

\[ \ln\bigl(1+t\bigr) \lt t \lt \bigl(1+t\bigr)\ln\bigl(1+t\bigr) \qquad (t\gt 0). \tag{2} \]

Each half of (2) is now a statement about a single elementary function.

Step 3: The left half

Let $\varphi(t) = t-\ln(1+t)$, so $\varphi(0) = 0$ and

\[ \varphi'(t) = 1-\frac{1}{1+t} = \frac{t}{1+t}\gt 0 \qquad (t\gt 0). \]

Hence $\varphi(t)\gt 0$, i.e. $\ln(1+t)\lt t$.

Step 4: The right half

Let $\psi(t) = \bigl(1+t\bigr)\ln(1+t)-t$, so $\psi(0) = 0$ and

\[ \psi'(t) = \ln(1+t)+\bigl(1+t\bigr)\cdot\frac{1}{1+t}-1 = \ln(1+t)\gt 0 \qquad (t\gt 0). \]

Hence $\psi(t)\gt 0$, i.e. $t\lt \bigl(1+t\bigr)\ln(1+t)$.

Step 5: Back to the original form

Both halves of (2) hold for every $t\gt 0$; substituting $t = \tfrac1x$ and multiplying by $x$ restores (1), and exponentiating gives the assertion.

Step 6: Verification

\[x = 1:\ 2 \lt e \lt 4; \qquad x = 10:\ 2.5937425 \lt 2.7182818 \lt 2.8531167 \;\;\checkmark\]

\[x = 100:\ 2.7048138 \lt e \lt 2.7318620 \;\;\checkmark\]

\[\text{Both sides tend to } e \text{ as } x\to\infty,\ \text{so the bracketing is asymptotically exact} \;\;\checkmark\]