Show that for $a,b>0$ and $n \in \mathbb N^+$ and $a \ne b$ $\left[a-\left(n+1\right)\left(a-b\right)\right]a^{n}Show that for $a,b>0$ and $n \in \mathbb N^+$ and $a \ne b$ the following does hold:
$$\left[a-\left(n+1\right)\left(a-b\right)\right]a^{n}<b^{n+1}$$

I tried to simplify the inequality to $$na^n(b-a)<b(b^n-a^n)$$
$$na^n(b-a)<b(b-a)(\sum_{k=0}^{n-1}b^{k}a^{n-1-k})$$
How to continue? (Notice that I don't know if $b-a$ is necessarily positive).
 A: It is equivalent to
$$a^{n+1}-b^{n+1}< (n+1)(a-b)a^{n}.$$
Assume $a>b$, then
$$\begin{aligned}
a^{n+1}-b^{n+1}&=(a-b)\cdot \sum_{i=0}^{n} a^i b^{n-i}\\
&< (a-b) (n+1)a^{n}.
\end{aligned}$$
And also
$$\begin{aligned}
a^{n+1}-b^{n+1}&=(a-b)\cdot \sum_{i=0}^{n} a^i b^{n-i}\\
&> (a-b) (n+1)b^{n},
\end{aligned}$$
that is
$$b^{n+1}-a^{b+1}< (b-a)(n+1)b^{n}.$$
A: We want to prove that, for any $n\in\mathbb{N}^+$ and $a,b>0;\;a\ne b$ the following inequality holds
$$a^n (a-(n+1) (a-b))-b^{n+1}<0$$
Proof by induction. It is true for $n=1$
$$a^1 (a-(1+1) (a-b))-b^{1+1}=-(a-b)^2<0$$
Now suppose it is true for $n$ that is
$$P(n)=-n a^{n+1}+b a^n+b n a^n-b^{n+1}<0\tag{1}$$
and let's prove it for $(n+1)$
$$P(n+1)=2 b a^{n+1}+b n a^{n+1}-n a^{n+2}-a^{n+2}-b^{n+2}$$
write the quotient and perform the long division wrt $b$
$$\frac{P(n+1)}{P(n)}=\frac{2 b a^{n+1}+b n a^{n+1}-n a^{n+2}-a^{n+2}-b^{n+2}}{-n a^{n+1}+b a^n+b n a^n-b^{n+1}}=\\=\frac{b(-n a^{n+1}+b a^n+b n a^n-b^{n+1})+(-2 b a^{n+1}-2 b n a^{n+1}+n a^{n+2}+a^{n+2}+b^2 a^n+b^2 n a^n)}{-n a^{n+1}+b a^n+b n a^n-b^{n+1}}$$
$$b+\frac{a^n \left(a^2 n+a^2-2 a b n-2 a b+b^2 n+b^2\right)}{n a^{n+1}-b n a^n-b a^n+b^{n+1}}=b+\frac{a^n (n+1) (a-b)^2}{n a^{n+1}-b n a^n-b a^n+b^{n+1}}$$
as the result is positive and the denominator is negative, the numerator must be negative, that is $P(n+1)<0$
