Say we have i.i.d. random variables $X_1,\cdots, X_n$ with the one-parameter Weibull distribution specified below: $$f(x\mid\lambda)=\lambda x^{\lambda-1}\exp(-x^{\lambda})$$ where $\lambda > 0$ and $x > 0$ with $\lambda$ is unknown. We wish to show there is a unique MLE.
I figured the likelihood function is:
$$L(\lambda\mid x_i)=\prod_{i=1}^n \lambda x_i^{\lambda-1}\exp(-x_i^{\lambda})$$
which would make the log-likelihood function:
$$\ell(\lambda\mid x_i) = n\log(\lambda) + (\lambda-1)\sum_{i=1}^{n}\log(x_{i}) - \sum_{i=1}^n x_i^{\lambda} $$
Our MLE will then satisfy the following equation (derivative of the above wrt $\lambda$):
$$ 0 = \frac{n}{\lambda} + \sum_{i=1}^n \log(x_i) - \sum_{i=1}^n \log(x_i)x_i^{\lambda}$$
This equation looks pretty unfriendly to solve in terms of $\lambda$. Is there something I'm missing somewhere earlier that can show the value of $\lambda$ must be unique?