This question is for all $a,m$ and $n$ $\in \mathbb{N}$
I'm really stuck on this question, I tried splitting the modulus and doing the congruence one at a time but can't find a way to put it back using CRT.
Because
$(a+1)^n \equiv 0\pmod {a+1}$ and $a^m \equiv 0\pmod a$
Also, I found out by trying different integers that
$(a+1)^n \equiv 1\pmod a$
Will this be useful for proving the question?
Hint $\rm\ LHS-RHS\equiv 0\,$ mod $a$ and mod $a\!+\!1$ so it is divisible by $\,a\,$ and $\,a\!+\!1\,$ thus also by their lcm = product = $\,a(a\!+\!1)\,$ since they are coprime. Note that mod $\ a\!+\!1\!:\ a\!+\!1\equiv 0\,\Rightarrow\, a\equiv -1.$
Or, using CRT verify that both sides are solutions of $\, x\equiv 1\pmod{a},\ x\equiv (-1)^m\pmod{a\!+\!1}\,$ hence, by CRT uniqueness, they are congruent $\!\pmod{a(a+1)}\,$
Quite frequently, uniqueness theorems provide powerful tools for proving equalities.