
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
If the derivative of the function $y=f(x)$, denoted as $y'=f'(x)$, is still a function of $x$, then the derivative of $y'=f'(x)$, denoted as $y''=f''(x)$, is called the second-order derivative of the function $y=f(x)$, and is represented as ${y^{(2)}}={f^{(2)}}(x)$. Similarly, the $(n-1)$th derivative of the function $y=f(x)$ is called the $n$th derivative of $y=f(x)$, denoted as ${y^{(n)}}={f^{(n)}}(x)$. When calculating the $n$th derivative of $y=\ln (x+1)$, it has been found that $y'=\dfrac{1}{x+1}, {y^{(2)}}=-\dfrac{1}{{(x+1)}^{2}}, {y^{(3)}}=-\dfrac{1\cdot 2}{{(x+1)}^{3}}, {y^{(4)}}=-\dfrac{1\cdot 2\cdot 3}{{(x+1)}^{4}},...,$ Based on the reasoning above, the $n$th derivative of the function $y=\ln (x+1)$ is

Present the answer in LaTex format: \boxed{Your answer}
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