Asymptotic expansion of (1 + 1/n)ⁿ

Exponent. With $u=\frac1n$ and $\ln(1+u)=u-\frac{u^2}2+\frac{u^3}3+o(u^3)$:
$$n\ln\left(1+\frac1n\right)=1-\frac1{2n}+\frac1{3n^2}+o\!\left(\frac1{n^2}\right).$$

Exponentiation. Let $v=-\frac1{2n}+\frac1{3n^2}$. Then $e^v=1+v+\frac{v^2}2+o\left(\frac1{n^2}\right)$ with $v^2=\frac1{4n^2}+o\left(\frac1{n^2}\right)$:
$$e^v=1-\frac1{2n}+\frac1{3n^2}+\frac1{8n^2}+o\!\left(\frac1{n^2}\right)=1-\frac1{2n}+\frac{11}{24n^2}+o\!\left(\frac1{n^2}\right).$$

Conclusion. Multiplying by $e$ gives the claimed expansion.

Check. For $n=100$: the expansion gives $2.7048138+\dots$: $e-\frac e{200}+\frac{11e}{240000}=2.718282-0.013591+0.000125=2.704816$, and $(1.01)^{100}=2.704814$ ✓.

Remark. In particular $\left(1+\frac1n\right)^n$ increases to $e$ with $e-u_n\sim\frac e{2n}$: a very slow convergence.