Prove that (1+x)^(1+x) = 1+x+x²+ dfrac12x³+o(x³) as x to 0

Prove that $(1+x)^{1+x} = 1+x+x^2+\dfrac12x^3+o(x^3)$ as $x\to0$.

Step 1: Write as an exponential

Let $u=(1+x)\ln(1+x)$, so $(1+x)^{1+x}=e^u$; expand $u$ to order $x^3$, then $e^u$ to the same order.

Step 2: Expand $u=(1+x)\ln(1+x)$

Multiplying $\ln(1+x)=x-\tfrac{x^2}2+\tfrac{x^3}3+o(x^3)$ by $(1+x)$ and collecting by degree: \[x^1{:}\,x;\quad x^2{:}\,-\tfrac{x^2}2+x^2=\tfrac{x^2}2;\quad x^3{:}\,\tfrac{x^3}3-\tfrac{x^3}2=-\tfrac{x^3}6. \tag{1}\]

\[ u = x+\frac{x^2}2-\frac{x^3}6+o(x^3). \tag{2} \]

Step 3: Expand $e^u=1+u+\tfrac{u^2}2+\tfrac{u^3}6+o(u^3)$

Since $u=x+O(x^2)$: $u^2=x^2+x^3+O(x^4)$ and $u^3=x^3+O(x^4)$, so

\[ \frac{u^2}2 = \frac{x^2}2+\frac{x^3}2+o(x^3), \qquad \frac{u^3}6 = \frac{x^3}6+o(x^3). \tag{3} \]

Step 4: Collect all terms

\[ x^0{:}\,1;\quad x^1{:}\,x;\quad x^2{:}\,\tfrac{x^2}2+\tfrac{x^2}2=x^2;\quad x^3{:}\,-\tfrac{x^3}6+\tfrac{x^3}2+\tfrac{x^3}6=\tfrac{x^3}2. \tag{4} \]

Step 5: Verification

\[x=0.05\text{: }(1.05)^{1.05}\approx1.0525646,\ 1+x+x^2+\tfrac12x^3\approx1.0525625,\ |{\rm error}|\approx2.1\times10^{-6}\text{, order }x^4\text{ as expected}\ \checkmark\]