Posterior of probability density function with $x^{-k}$ and uniform prior

Question

Suppose you are sampling from a probability distribution which contains a reciprocal:

$$f(x; \alpha, \lambda) = \begin{cases}\frac{\lambda \alpha^\lambda}{x^{\lambda+1}}~~x\geq \alpha;\\ 0 ~~~~~~\text{otherwise};\end{cases}$$

As explained here, the likelihood is maximised at $=min(\mathbf X)$.

The prior is assumed to be uniform $U[0,a] $

The posterior is proportional to the likelihood multiplied by the prior :

$$ \pi(\alpha|\mathbf x) \propto \frac{1}{a} \frac{\lambda^n \alpha^{n\lambda}} {\prod x_i^{\lambda+1}} ~~~~~~~~~ 0<\alpha<min(\mathbf X) $$

The integrand must be equal to 1:

$$ \int_0^{min(\mathbf x)} \frac{k}{a} \frac{\lambda^n \alpha^{n\lambda}} {\prod x_i^{\lambda+1}} d\alpha= 1 $$

$$ k = \frac{a\prod x_i^{\lambda+1}(n\lambda+1)}{\lambda^nmin(\mathbf x)^{n\lambda+1}} $$

Plugging this back into the posterior I get:

$$ \pi(\alpha|\mathbf x) = \frac{(n\lambda+1)\alpha^{n\lambda}}{min(\mathbf x)^{n\lambda+1}} $$

However I was expecting the posterior to be dependent on $a$, as well as $min(\mathbf x)$. Any suggestions on where I have gone wrong would be much appreciated!