I cannot speak as to the use of these symbols but let me show you instead the traditional way, why the mle is biased.
Recall that the exponential distribution is a special case of the General Gamma distribution with two parameters, shape $a$ and rate $b$. The pdf of a Gamma Random Variable is:
$$f_Y (y)= \frac{1}{\Gamma(a) b^a} y^{a-1} e^{-y/b}, \ 0<y<\infty$$
where $\Gamma (.)$ is the gamma function. Alternative parameterisations exist, see for example the wikipedia page.
If you put $a=1$ and $b=1/\lambda$ you arrive at the pdf of the exponential distribution:
$$f_Y(y)=\lambda e^{-\lambda y},0<y<\infty$$
One of the most important properties of a gamma RV is the additivity property, simply put that means that if $X$ is a $\Gamma(a,b)$ RV, $\sum_{i=1}^n X_i$ is also a Gamma RV with $a^{*}=\sum a_i$ and $b^{*}=b$ as before.
Define $Y=\sum X_i$ and as noted above $Y$ is also a Gamma RV with shape parameter equal to $n$, $\sum_{i=1}^n 1 $, that is and rate parameter $1/\lambda$ as $X$ above. Now take the expectation $E[Y^{-1}]$
$$ E\left [ Y^{-1} \right]=\int_0^{\infty}\frac{y^{-1}y^{n-1}\lambda^n}{\Gamma(n)}\times e^{-\lambda y}dy=\int_0^{\infty}\frac{y^{n-2}\lambda^n}{\Gamma(n)}\times e^{-\lambda y}dy$$
Comparing the latter integral with an integral of a Gamma distribution with shape parameter $n-1$ and rate one $1/\lambda$ and using the fact that $\Gamma(n)=(n-1) \times \Gamma(n-1)$ we see that it equals $\frac{\lambda}{n-1}$. Thus
$$E\left[ \hat{\theta} \right]=E\left[ \frac{n}{Y} \right]=n \times E\left[Y^{-1}\right]=\frac{n}{n-1} \lambda$$
which clearly shows that the mle is biased. Note, however, that the mle is consistent. We also know that under some regularity conditions, the mle is asymptotically efficient and normally distributed, with mean the true parameter $\theta$ and variance $\{nI(\theta) \}^{-1} $. It is therefore an optimal estimator.
Does that help?
In a reply concerning interpreting probability densities I have argued for the merits of including the differential terms ($dx$ and $dy$ in this case) in the PDF, so let's write
$$f(x,y; \theta,\lambda) = \frac{1}{\lambda x \sqrt{2\pi}} \exp\left(-\frac{(y-x\theta)^2}{2x^2} - \frac{x}{\lambda}\right)\,dx\,dy.$$
In (d), attention is focused on $\hat{\theta}$ which appears to be a multiple of $\sum_i y_i/x_i$ where each $(x_i,y_i)$ is an independent realization of the bivariate random variable described by $f$. To tackle this, let's consider the distribution of $Z=Y/X$--and then we will later have to sum a sequence of independent versions of this variable in order to determine the distribution of $\hat\theta$.
In contemplating the change of variable $(X,Y)\to (X,Z) = (X,Y/X)$, we recognize that the non-negativity of $X$ assures this induces an order-preserving, one-to-one correspondence between $Y$ and $Z$ for each $X$. Therefore all we need to do is change the variable in the integrand, writing $y=x z$:
$$f(x,x z; \theta,\lambda) = \frac{1}{\lambda x \sqrt{2\pi}} \exp\left(-\frac{(x z-x\theta)^2}{2x^2} - \frac{x}{\lambda}\right)\,dx\,d(x z).$$
From the product rule $d(x z) = z dx + x dz$, and understanding the combination $dx\, d(x z)$ in the sense of differential forms $dx \wedge d(x z)$, we mechanically obtain
$$dx\, d(x z) = dx \wedge d(x z) = dx \wedge (z dx + x dz) = z dx \wedge dx + x dx \wedge dz = x\,dx\,dz.$$
(This is the easy way to obtain the Jacobian determinant of the transformation.)
Therefore the distribution of $(X,Z)$ has PDF
$$\eqalign{
g(x,z;\theta,\lambda) = f(x, x z; \theta, \lambda) &= \frac{1}{\lambda x \sqrt{2\pi}} \exp\left(-\frac{(x z-x\theta)^2}{2x^2} - \frac{x}{\lambda}\right)\,x dx\,dz \\
&= \frac{1}{\lambda} \exp\left( - \frac{x}{\lambda}\right)\frac{1}{\sqrt{2\pi}}\,dx\ \exp\left(-\frac{( z-\theta)^2}{2}\right) \,dz.
}$$
Because this has separated into a product of a PDF for $X$ and a PDF for $Z$, we see without any further effort that (1) $X$ and $Z$ are independent and (2) $Z$ has a Normal$(\theta,1)$ distribution.
Finding the joint distribution of $(\hat{\lambda}, \hat\theta)$, which are easily expressed in terms of $X$ and $Z$, is now straightforward.
Best Answer
The MLE and likelihood function are invariant for bijective functions of a parameter.
$$P\left[x \:\vert\: f(\theta)=f(a)\right] = P\left[x \:\vert\: \theta=a\right]$$
Only when the parameter can have negative values there might be a difference between the MLE of parameter and the square of a parameter. (because two values map to the same square $x \mapsto x^2$ and also $-x \mapsto x^2$)
So $(\lambda^2)_{mle}=(\lambda_{mle})^2$