I have a very simple case: Find general formula for $\frac{a^{2n+1}-a^{2n-1}}{a^n-a^{n-1}}$. Of course dividing one by another was quite simple with outcome: $a^n(a+1)$. However I would like to prove it. My question is: is this enough as a proof? If this was something really complicated, and I would prove it in such manner, would that be enough?
for $n=1$:
$$\frac{a^3-a}{a-1}=a^2+a=a(a+1)$$
for $n=2$:
$$\frac{a^5-a^3}{a^2-a}=a^3+a^2=a^2(a+1)$$
for $N=n+1$:
$$\frac{a^{2(n+1)+1}-a^{2(n+1)-1}}{a^{n+1}-a^{(n+1)-1}}=\frac{a^{2n+3}-a^{2n+1}}{a^{n+1}-a^n}=\frac{a^2(a^{2n+1}-a^{2n-1})}{a(a^n-a^{n-1})}=a\cdot a^n(a+1)=a^{n+1}(a+1)$$
$$\frac{a^{n+1}-a^{n-1}}{a^n-a^{n-1}}$$
$$=\frac{a^{n-1}(a^2-1)}{a^{n-1}(a-1)}$$
$$=a+1$$ assuming $a(a-1)\ne0$