Since you are a tutor, any knowledge is always for a good cause. So I will provide some bounds for the MLE.
We have arrived at
$$(1-\lambda x_{(n)})e^{\lambda x_{(n)} } + \lambda n x_{(n)} - 1 = 0$$
with $x_{(n)}\equiv M_n$. So
$$(1-\hat \lambda x_{(n)})e^{\hat \lambda x_{(n)}} = 1-\hat \lambda x_{(n)}n $$
Assume first that $1-\hat \lambda x_{(n)} >0$. Then we must also have $1-\hat \lambda x_{(n)}n>0$ since the exponential is always positive. Moreover since $x_{(n)}, \hat \lambda > 0\Rightarrow e^{\hat \lambda x_{(n)}}>1$. Therefore we should have
$$\frac {1-\hat \lambda x_{(n)}n}{1-\hat \lambda x_{(n)}}>1 \Rightarrow \hat \lambda x_{(n)}>\hat \lambda x_{(n)}n$$
which is impossible. Therefore we conclude that
$$\hat \lambda >\frac 1{x_{(n)}},\;\; \hat \lambda = \frac c{x_{(n)}}, \;\; c>1$$
Inserting into the log-likelihood we get
$$\ell(\hat\lambda(c)\mid x_{(n)}) = \log \frac c{x_{(n)}} + \log n - \frac c{x_{(n)}} x_{(n)} + (n-1) \log (1 - e^{-\frac c{x_{(n)}} x_{(n)}})$$
$$= \log \frac n{x_{(n)}} + \log c - c + (n-1) \log (1 - e^{-c})$$
We want to maximize this likelihood with respect to $c$. Its 1st derivative is
$$\frac{d\ell}{dc}=\frac 1c -1 +(n-1)\frac 1{e^{c}-1}$$
Setting this equal to zero, we require that
$$e^{c}-1 - c\left(e^{c}-1\right)+(n-1)c =0$$
$$\Rightarrow \left(n-e^c\right)c = 1-e^c$$
Since $c>1$ the RHS is negative. Therefore we must also have $n-e^c <0 \Rightarrow c > \ln n$. For $n\ge 3$ this provides a tighter lower bound for the MLE, but it doesn't cover the $n=2$ case, so
$$\hat \lambda > \max \left\{\frac 1{x_{(n)}}, \frac {\ln n}{x_{(n)}}\right\}$$
Moreover (for $n\ge 3$) rearranging the 1st-order condition we have that
$$c= \frac{e^c-1}{e^c-n} > \ln n \Rightarrow e^c -1 > e^c\ln n -n\ln n $$
$$\Rightarrow n\ln n-1>e^c(\ln n -1) \Rightarrow c< \ln{\left[\frac{n\ln n-1}{\ln n -1}\right]}$$
So for $n\ge 3$ we have that
$$\frac 1{x_{(n)}}\ln n < \hat \lambda < \frac 1{x_{(n)}}\ln{\left[\frac{n\ln n-1}{\ln n -1}\right]}$$
This is a narrow interval, especially if $x_{(n)}\ge 1$. For example (truncated at 3d digit )
$$\begin{align}
n=10 & &\frac 1{x_{(n)}}2.302 < \hat \lambda < \frac 1{x_{(n)}}2.827\\
n=100 & & \frac 1{x_{(n)}}4.605 < \hat \lambda < \frac 1{x_{(n)}}4.847\\
n=1000 & & \frac 1{x_{(n)}}6.907 < \hat \lambda < \frac 1{x_{(n)}}7.063\\
n=10000 & & \frac 1{x_{(n)}}9.210< \hat \lambda < \frac 1{x_{(n)}}9.325\\
\end{align}$$
Numerical examples indicate that the MLE tends to be equal to the upper bound, up to second decimal digit.
ADDENDUM: A CLOSED FORM EXPRESSION
This is just an approximate solution (it only approximately maximizes the likelihood), but here it is:
manipulating the 1st-order condition we want to have
$$\lambda = \frac 1{x_{(n)}}\ln \left[\frac {\lambda x_{(n)}n -1}{\lambda x_{(n)} -1}\right]$$
Now, one can show (see for example here) that
$$E[X_{(n)}] = \frac {H_n}{\lambda},\;\; H_n = \sum_{k=1}^n\frac 1k$$
Solving for $\lambda$ and inserting into the RHS of the implicit 1st-order condition, we obtain
$$\lambda = \frac 1{x_{(n)}}\ln \left[\frac {nH_n\frac {x_{(n)}}{E[X_{(n)}]} -1}{ H_n\frac {x_{(n)}}{E[X_{(n)}]} -1}\right]$$
We want an estimate of $\lambda$, given that $X_{(n)}=x_{(n)}$, $\hat \lambda \mid \{X_{(n)}=x_{(n)}\}$. But in such a case, we also have $E[X_{(n)}\mid \{X_{(n)}=x_{(n)}\}] =x_{(n)}$. this simplifies the expression and we obtain
$$\hat \lambda = \frac 1{x_{(n)}}\ln \left[\frac {nH_n -1}{ H_n -1}\right]$$
One can verify that this closed form expression stays close to the upper bound derived previously, but a bit less than the actual (numerically obtained) MLE.
Best Answer
As a number The maximum of a sample is the highest number of a sample. This is a single number.
As a function But, when we consider the sample as a random variable that can take different values with different probabilities, then the maximum becomes a random variable that can take different values with different probabilities.
In this case we can describe the maximum with a function that described this probability for a specific value. (and the same for other order statistics)
Example: say you roll two six sided dices and you consider the highest value.
For and particular roll the maximum is the highest number of the roll. For example, if you roll '4' and '1', then the maximum is '4'.
For the distribution of the possible rolls of the maximum we compute the probabilities for a particular maximum. This will be the probability as function of the value of the maximum.
The figure below shows the maximum as function of the two dice rolls
$$ \begin{array}{c|cccccc} &\color{red}1 & \color{orange}2 & \color{gold} 3 & \color{green} 4 & \color{blue}5 &\color{purple} 6 \\ \hline \color{red}1&\color{red} 1 & \color{orange}2 & \color{gold} 3 & \color{green}4 & \color{blue}5 &\color{purple} 6\\ \color{orange}2&\color{orange}2 &\color{orange} 2 & \color{gold} 3 & \color{green}4 & \color{blue}5 &\color{purple} 6\\ \color{gold} 3&\color{gold} 3 & \color{gold} 3 & \color{gold} 3 &\color{green} 4 & \color{blue}5 & \color{purple}6\\ \color{green}4&\color{green}4 & \color{green}4 & \color{green}4 &\color{green} 4 &\color{blue} 5 & \color{purple}6\\ \color{blue}5&\color{blue}5 &\color{blue} 5 &\color{blue} 5 &\color{blue} 5 &\color{blue} 5 &\color{purple} 6\\ \color{purple}6&\color{purple}6 &\color{purple} 6 & \color{purple}6 &\color{purple} 6 & \color{purple}6 & \color{purple}6\\ \end{array}$$
Then you can see that the probability that the maximum is $1$ is $\frac{1}{36}$, the probability that the maximum is $2$ is $\frac{3}{36}$, the probability that the maximum is $3$ is $\frac{5}{36}$, and so on.
We can describe it as a function
$$ \mathbb{P}(\text{max(2 six sided dice rolls)} = k) = \frac{2k-1}{36}$$