Problem on limit of sequences 
I started solving the question by taking  $b_{n}=\frac{(1^{1^{^{p}}}2^{2^{p}}...(n+1){^{(n+1)^{p}}})^{\frac{1}{(n+1)^{p+1}}}}{  (1^{1^{^{p}}}2^{2^{p}}...(n){^{(n)^{p}}})^{\frac{1}{(n)^{p+1}}}}$
 where  $\lim _{n \to \infty }b_{n}=e^{\frac{-2}{(p+1)^{2}}}$ and
$lim_{n \to \infty } x_{n}=(1^{1^{^{p}}}2^{2^{p}}...(n){^{(n)^{p}}})^{\frac{1}{(n)^{p+1}}}(b_{n}-1)$ . But I don't know how to solve it further can someone please  help  me with a hint or suggest some other method for solving this question  
 A: Let $x_n=y_{n+1}-y_n$ where
$$
y_n=\left(1^{1^p}\cdot 2^{2^p} \cdots n^{n^p}\right)^{\frac1{n^{p+1}}}.
$$
Then 
$$
\begin{align}\ln y_n&=\frac1{n^{p+1}}\sum_{i=1}^n i^p \ln i=\frac1n\left(\sum_{i=1}^n (\frac in)^p \left( \ln \frac in +  \ln n\right)\right)\\
&=\frac1n \sum_{i=1}^n (\frac in)^p \ln \frac in + \left(\frac1n \sum_{i=1}^n (\frac in)^p \right)\ln n\\
&=\int_0^1 x^p \ln x dx + O(\frac1n) + \left( \int_0^1 x^p dx + O(\frac 1n)\right) \ln n\\
&=\frac{x^{p+1}}{p+1}\ln x \bigg\vert_0^1-\int_0^1 \frac{x^p}{p+1} dx +\frac{\ln n}{p+1} +O(\frac{\ln n}{n})\\
&=-\frac1{(p+1)^2}+\ln n^{\frac1{p+1}}+O(\frac{\ln n}n)
\end{align}
$$
Thus, 
$$
y_n=e^{-\frac1{(p+1)^2}}n^{\frac1{p+1}}\left(1+O(\frac{\ln n}n)\right)=e^{-\frac1{(p+1)^2}}n^{\frac1{p+1}}+O(\frac{\ln n}{n^{\frac p{p+1}}})
$$
For $p>0$, we have by Mean Value Theorem, there is $c_n\in (n,n+1)$ such that
$$
y_{n+1}-y_n = e^{-\frac1{(p+1)^2}} \frac1{p+1} c_n^{-\frac p{p+1}}+O(\frac{\ln n}{n^{\frac p{p+1}}})\xrightarrow[n\rightarrow\infty]{} 0.
$$
For $p=0$, 
$$
\ln y_n=\frac1n \sum_{i=1}^n \ln i=\frac1n (n\ln n - n + \frac12\ln (2\pi n)+O(\frac1n))=\ln n -1 +\frac{\ln (2\pi n)}{2n}+O(\frac1{n^2}).
$$
This gives
$$
y_n= e^{-1} n (2\pi n)^{\frac1{2n}}\left(1+O(\frac{1}{n^2})\right)=e^{-1} n (2\pi n)^{\frac1{2n}}+O(\frac1n).
$$
Therefore, by Mean Value Theorem, there is $c_n\in (n,n+1)$ such that
$$
y_{n+1}-y_n = e^{-1} \left((2\pi c_n)^{\frac1{2c_n}}+c_n (2\pi c_n)^{\frac1{2c_n}}\frac{2-2\ln(2\pi c_n) }{4c_n^2}\right)+O(\frac1n)\xrightarrow[n\rightarrow\infty]{} e^{-1}.
$$
