Let $ a>0$, $ d>0$ and put 
$ f(x)\equal{}\frac{1}{a}\plus{}\frac{x}{a(a\plus{}d)}\plus{}\cdots\plus{}\frac{x^n}{a(a\plus{}d)\cdots(a\plus{}nd)}\plus{}\cdots$
Give a closed form for $ f(x)$.
1. **Define the function \( f(x) \):**
   \[
   f(x) = \sum_{n=0}^{\infty} \frac{x^n}{a(a+d)(a+2d)\cdots(a+nd)}
   \]

2. **Introduce the function \( g(x) \):**
   \[
   g(x) = x^a f(x^d)
   \]

3. **Differentiate \( g(x) \):**
   \[
   g'(x) = \frac{d}{dx} \left( x^a f(x^d) \right)
   \]
   Using the product rule:
   \[
   g'(x) = a x^{a-1} f(x^d) + x^a \frac{d}{dx} f(x^d)
   \]
   Since \( f(x^d) \) is a function of \( x^d \), we use the chain rule:
   \[
   \frac{d}{dx} f(x^d) = f'(x^d) \cdot d x^{d-1}
   \]
   Therefore:
   \[
   g'(x) = a x^{a-1} f(x^d) + x^a \cdot d x^{d-1} f'(x^d)
   \]
   Simplifying:
   \[
   g'(x) = a x^{a-1} f(x^d) + d x^{a+d-1} f'(x^d)
   \]

4. **Express \( g'(x) \) in terms of \( g(x) \):**
   \[
   g'(x) = x^{a-1} + x^{d-1} g(x)
   \]
   This is given in the problem statement.

5. **Solve the differential equation:**
   \[
   g'(x) = x^{a-1} + x^{d-1} g(x)
   \]
   This is a first-order linear differential equation. We can solve it using an integrating factor. The standard form is:
   \[
   g'(x) - x^{d-1} g(x) = x^{a-1}
   \]
   The integrating factor \( \mu(x) \) is:
   \[
   \mu(x) = e^{\int -x^{d-1} \, dx}
   \]
   For \( d \neq 1 \):
   \[
   \mu(x) = e^{-\frac{x^d}{d}}
   \]
   Multiply both sides of the differential equation by \( \mu(x) \):
   \[
   e^{-\frac{x^d}{d}} g'(x) - e^{-\frac{x^d}{d}} x^{d-1} g(x) = e^{-\frac{x^d}{d}} x^{a-1}
   \]
   The left-hand side is the derivative of \( g(x) e^{-\frac{x^d}{d}} \):
   \[
   \frac{d}{dx} \left( g(x) e^{-\frac{x^d}{d}} \right) = e^{-\frac{x^d}{d}} x^{a-1}
   \]
   Integrate both sides:
   \[
   g(x) e^{-\frac{x^d}{d}} = \int e^{-\frac{x^d}{d}} x^{a-1} \, dx + C
   \]
   For \( d = 1 \):
   \[
   \mu(x) = e^{-\int x^{0} \, dx} = e^{-x}
   \]
   Multiply both sides of the differential equation by \( e^{-x} \):
   \[
   e^{-x} g'(x) - e^{-x} g(x) = e^{-x} x^{a-1}
   \]
   The left-hand side is the derivative of \( g(x) e^{-x} \):
   \[
   \frac{d}{dx} \left( g(x) e^{-x} \right) = e^{-x} x^{a-1}
   \]
   Integrate both sides:
   \[
   g(x) e^{-x} = \int e^{-x} x^{a-1} \, dx + C
   \]

6. **Solve the integral:**
   For \( d \neq 1 \):
   \[
   g(x) = e^{\frac{x^d}{d}} \left( \int e^{-\frac{x^d}{d}} x^{a-1} \, dx + C \right)
   \]
   For \( d = 1 \):
   \[
   g(x) = e^{x} \left( \int e^{-x} x^{a-1} \, dx + C \right)
   \]

7. **Find the closed form of \( f(x) \):**
   Since \( g(x) = x^a f(x^d) \), we need to express \( f(x) \) in terms of \( g(x) \):
   \[
   f(x^d) = \frac{g(x)}{x^a}
   \]
   Therefore:
   \[
   f(x) = \frac{g(x^{1/d})}{(x^{1/d})^a} = \frac{g(x^{1/d})}{x^{a/d}}
   \]

The final answer is \( \boxed{ f(x) = \frac{g(x^{1/d})}{x^{a/d}} } \).