Evaluate $f^{2016}(0)$ for the function $f(x)=x^2 \ln(x+1)$ 

Question: Evaluate $f^{2016}(0)$ for the function $f(x)=x^2 \ln(x+1)$ 


Background: I know we can use taylor/mclaurin/power series to solve this but these were not teached to me by my teacher and not part of the syllabus so we are required to find the $nth$ derivative and make a suitable observation and hence evaluate the function. 
One thing I wasn't sure about was to make a new post or edit my old post asking the part of the question. So if someone could clarify that for me , it would be great for the future. 

My attempt (for $n\geq 3$ I have used partial fractions decomposition but I have not showed the working):
$$f^{(0)}(x)=x^2 \ln(x+1)$$
$$f^{(1)}(x)= 2x\ln(x+1) + \frac{x^2}{x+1} $$
$$f^{(2)}(x)=2\ln(x+1) + \frac{4x}{x+1} - \frac{x^2}{(x+1)^2}$$
$$f^{(3)}(x)=\frac{2}{(x+1)}+\frac{2}{(x+1)^2}+\frac{2}{(x+1)^3}$$
$$f^{(4)}(x)=-\frac{2}{(x+1)^2}-\frac{6}{(x+1)^3}-\frac{8}{(x+1)^4} $$
$$f^{(5)}(x)=\frac{4}{(x+1)^3}+\frac{12}{(x+1)^4}+\frac{24}{(x+1)^5}$$
From here I can see that for $n\geq3$ we have: 
$$ f^{(n)}(x)=(-1)^{n-1} \left[ \frac{2(n-3)!}{(x+1)^{n-2}} + \frac{2(n-2)!}{(x+1)^{n-1}} + \frac{(n-1)!}{(x+1)^n} \right]$$
(it doesn't work for $n <3$ since we can't have a negative factorial)
Now evaluating this at zero for the $2016$th derivative:
$$ f^{(2016)}(0)=(-1)^{2015} \left[ 2(2013)!+2(2014)!+2015! \right]$$
$$ \Longrightarrow f^{(2016)}(0)=(-1) \left[ 2(2013)!+2(2014)!+2015! \right]$$
Would this be correct?
 A: It is okay to use partial fractions and derive the result, but you should still eventually prove that $n$-th derivative formula is correct. You claim it is for $n\geq 3$, so why not to use induction for this?
For $n=3$ you can simply verify this (and you did that).
For induction step, you assume 
$$f^{(n)}(x)=(-1)^{n-1} \left[ \frac{2(n-3)!}{(x+1)^{n-2}} + \frac{2(n-2)!}{(x+1)^{n-1}} + \frac{(n-1)!}{(x+1)^n} \right]$$
and you want to show
$$f^{(n+1)}(x)=(-1)^{n+1-1} \left[ \frac{2(n+1-3)!}{(x+1)^{n+1-2}} + \frac{2(n+1-2)!}{(x+1)^{n+1-1}} + \frac{(n+1-1)!}{(x+1)^{n+1}} \right]=(-1)^{n} \left[ \frac{2(n-2)!}{(x+1)^{n-1}} + \frac{2(n-1)!}{(x+1)^{n}} + \frac{n!}{(x+1)^{n+1}} \right]$$
But you have
\begin{align}
f^{(n+1)}(x) &= \frac{d}{dx} f^{(n)}(x) \\
&= \frac{d}{dx} \left((-1)^{n-1} \left[ \frac{2(n-3)!}{(x+1)^{n-2}} + \frac{2(n-2)!}{(x+1)^{n-1}} + \frac{(n-1)!}{(x+1)^n} \right]\right) \\
&= \left((-1)^{n-1} \left[ \frac{d}{dx}\frac{2(n-3)!}{(x+1)^{n-2}} + \frac{d}{dx}\frac{2(n-2)!}{(x+1)^{n-1}} + \frac{d}{dx}\frac{(n-1)!}{(x+1)^n} \right]\right) \\
\end{align}
Now just derive the fractions and after simplification you will obtain the expected result for $f^{(n+1)}(x)$.
After you prove this you can safely use the formula to evaluate $f^{(2016)}(0)$ the way you suggested.
A: We can use the General Leibniz rule, that is: $$(uv)^{(n)}=\sum_{k=0}^n\binom{n}{k}u^{(n-k)}v^{(k)}$$ In our case, taking $u(x)=\ln (x+1)$ and $v(x)=x^2$: 
$$\left \{ \begin{matrix} u^{(0)}(x)=\log (1+x) \\ u^{(1)}(x)=(x+1)^{-1}\\ u^{(2)}(x)=-1(x+1)^{-2}\\u^{(3)}(x)=2(x+1)^{-3}\\ \ldots \\ u^{(n)}(x)=(-1)^{n+1}(n-1)!(x+1)^{-n}\end{matrix}\right.\;\;\; \left \{ \begin{matrix} v^{(0)}(x)=x^2 \\ v^{(1)}(x)=2x \\ v^{(2)}(x)=2\\ v^{(3)}(x)=0 \\ \ldots \\ v^{(n)}(x)=0\end{matrix}\right.$$ 
$$\left \{ \begin{matrix} u^{(0)}(0)=0 \\ u^{(1)}(0)=1\\ u^{(2)}(0)=-1\\u^{(3)}(0)=2\\ \ldots \\ u^{(n)}(0)=(-1)^{n+1}(n-1)!\end{matrix}\right.\;\;\; \left \{ \begin{matrix} v^{(0)}(0)=0 \\ v^{(1)}(0)=0 \\ v^{(2)}(0)=2\\ v^{(3)}(0)=0 \\ \ldots \\ v^{(n)}(0)=0\end{matrix}\right.$$
 $$f^{(2016)}(0)=(uv)^{(2016)}(0)=\displaystyle\binom{2016}{2}u^{(2014)}(0)v^{(2)}(0)$$ $$=\displaystyle\binom{2016}{2}(-1)(2013)!\cdot 2=\ldots$$
