Find the next year of the year $2016$ written as $n^{m-1}(n^{m}-1)$ See that : $2016=2^{6-1}(2^{6}-1)$ 
Find the next year of the year $2016$ written as $n^{m-1}(n^{m}-1)$ where $n,m>1$ natural numbers
I don't have any ideas to start in this problem
My be we need the factorization of $2016=2^{5}3^{2}.7$
But I need know the relation between factorization primary and second

Since $n^{m-1}(n^{m}-1)$ increases quite fast ,  we can try inputting some values of $m$ 

For $m=2$ , we get $n(n^2-1) \gt 2016$   , which is true for $n \ge 13. $ So the smallest value is at $n = 13$ , which gives $13 \times 168 = \color{#2dd}{2184 }.$

For $m=3$ , we get $n^2(n^3-1) \gt 2016$ , which is true for $n \ge 5 .$ So the smallest value is at $n = 5$ , which gives $25 \times 124 = \color{#5d0}{3100}.$

For $m=4$ , we get $n^3(n^4-1) \gt 2016$ , which is true for $n \ge 3 .$ So the smallest value is at $n = 3$ , which gives $27\times 80 = \color{#d05}{2160} . $

For $m=5$ , we get $n^4(n^5-1) \gt 2016$ , which is true for $n \ge 3 .$ So the smallest value is at $n = 3$ , which gives $81\times 242 = \color{#20f}{19602 }.$

For $m=6$ , we already have the value $2^5(2^6-1)$ and for $m \gt 6$ , the equation is not true for $n,m \gt 1$
So the next smallest value is $\boxed{2160 }.$
