Use power series operations to find the Taylor series at $x=0$ for the given function.
\[
f(x)=x^{7} e^{x}
\]
A) $\sum_{n=0}^{\infty} \frac{x^{n+7}}{(m+7) !}$
B) $\sum_{n=0}^{\infty} \frac{7^{n} x^{n}}{n !}$
C) $\sum_{n=0}^{\infty} \frac{7 n x^{n}}{(n+7) !}$
D) $\sum_{n=0}^{\infty} \frac{x^{n+7}}{n !}$

Step 1 ：The Taylor series expansion of a function \(f(x)\) about \(x=0\) is given by: \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\] where \(f^{(n)}(0)\) is the nth derivative of \(f(x)\) evaluated at \(x=0\).

Step 2 ：The function given is \(f(x) = x^7 e^x\). We know that the Taylor series expansion for \(e^x\) is \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\).

Step 3 ：We can use this to find the Taylor series for \(f(x) = x^7 e^x\) by multiplying the series for \(e^x\) by \(x^7\). This gives us: \[f(x) = x^7 e^x = x^7 \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{x^{n+7}}{n!}\]

Step 4 ：So, the correct answer is D) \(\sum_{n=0}^{\infty} \frac{x^{n+7}}{n!}\).

Step 5 ：Final Answer: The Taylor series at \(x=0\) for the function \(f(x)=x^{7} e^{x}\) is \(\boxed{D) \sum_{n=0}^{\infty} \frac{x^{n+7}}{n!}}\).