Logs of this year

$\begin{aligned} \log_{n} \log_{n-1} \cdots \log_3 \log_2 2^{3^{{.}^{{.}^{{.}^{{n-1}^{n}}}}}} & = \log_{n} \log_{n-1} \cdots \log_4 \log_3 3^{4^{{.}^{{.}^{{.}^{{n-1}^{n}}}}}} \log_2 2\\ \\ & = \log_{n} \log_{n-1} \cdots \log_5\log_4 4^{5^{{.}^{{.}^{{.}^{{n-1}^{n}}}}}} \log_3 3\\ \\ & = \log_{n} \log_{n-1} \cdots \log_6 \log_5 5^{6^{{.}^{{.}^{{.}^{{n-1}^{n}}}}}} \log_4 4\\ \\ & \vdots\\ \\ & = \log_{n} \log_{n-1} {(n-1)}^{n}\\ \\ & = \log_{n} n \log_{n-1} n-1\\ \\ & = \color{#20A900}{1} \end{aligned}$

$\begin{aligned} \log_2 \log_3 \cdots \log_{n-1} \log_{n} {n}^{{n-1}^{{.}^{{.}^{{.}^{{3}^{2}}}}}} & = \log_2 \log_3 \cdots \log_{n-2} \log_{n-1} {n-1}^{{n-2}^{{.}^{{.}^{{.}^{{3}^{2}}}}}} \log_{n} n\\ \\ & = \log_2 \log_3 \cdots \log_{n-3} \log_{n-2} {n-2}^{{n-3}^{{.}^{{.}^{{.}^{{3}^{2}}}}}} \log_{n-1} n-1\\ \\ & = \log_2 \log_3 \cdots \log_{n-4} \log_{n-3} {n-3}^{{n-4}^{{.}^{{.}^{{.}^{{3}^{2}}}}}} \log_{n-2} n-2\\ \\ & \vdots\\ \\ & = \log_2 \log_3 3^2\\ \\ & = \log_2 2 \log_3 3\\ \\ & = \color{#20A900}{1} \end{aligned}$

So, even if we substitute $n = 2016$ , $\color{#3D99F6}{\boxed{\text{Both are equal}}}$ .