let us replace n with x, not neccessary, but might help to "see" the answer.
lim x-->oo [x[x+1] lal^[x+1]
if a is less than 1 than we can replace it with one over b, where b is a integer greater than 1
lim x-->oo [x[x+1] / lbl^[x+1] = oo/oo undefined use L'Hopitals rule
lim x-->oo [x+x+1] / lnb lbl^[x+1] = oo/oo
undefined use L'Hopitals rule
lim x-->oo 2/ [ln lbl ^2 l b l ^[x+1]] = 2/oo
lim goes to 0
whoever gave you the answer was right
I assume you have studied L'hopitals rule the derivstive over the derivative
Arthur