Find the minimum value of the sum

We can use Jensen's inequality to find the minimum value of $a_1^{n+1} + a_2^{n+1} + a_3^{n+1} + \cdots + a_n^{n+1}$ . Since $f(x)=x^{n+1}$ is convex, then

$\begin{aligned} \frac {w_1f(a_1)+w_2f(a_2)+w_3f(a_3)+\cdots + w_nf(a_n)}{w_1+w_2+w_3+\cdots+w_n} & \ge f \left(\frac {w_1a_1+w_2a_2+w_3a_3+\cdots + w_na_n}{w_1+w_2+w_3+\cdots+w_n} \right) \\ \frac {a_1^{n+1} + a_2^{n+1} + a_3^{n+1} + \cdots + a_n^{n+1}}{\underbrace{1+1+1+\cdots + 1}_{=n}} & \ge \left( \frac {a_1 + a_2 + a_3 + \cdots + a_n}{\underbrace{1+1+1+\cdots + 1}_{=n}} \right)^{n+1} = \left(\frac 1n \right)^{n+1} \\ \implies a_1^{n+1} + a_2^{n+1} + a_3^{n+1} + \cdots + a_n^{n+1} & \ge \left(\frac 1n \right)^n \end{aligned}$

Therefore, $\alpha = n \implies (\alpha - n)^{n+1} = \boxed 0$ .