Take an integer $k\geq0$, so $k\rho$ is a <dominant integral weight>. Apply the dilation $e^\mu\mapsto e^{(k+1)\mu}$ to the <Weyl denominator formula>:
$$
\sum_{w\in W}\det(w)e^{(k+1)w\rho}
=e^{(k+1)\rho}\prod_{\alpha\in R^+}(1-e^{-(k+1)\alpha}).
$$
This is exactly the numerator of the <Weyl character formula> for $L(k\rho)$. Divide by its denominator and use a finite <geometric series> in each root direction:
$$
\boxed{\operatorname{ch}L(k\rho)
=e^{k\rho}\prod_{\alpha\in R^+}
\frac{1-e^{-(k+1)\alpha}}{1-e^{-\alpha}}
=e^{k\rho}\prod_{\alpha\in R^+}(1+e^{-\alpha}+\cdots+e^{-k\alpha}).}
$$
This <character of a Weyl-vector multiple> also gives $\dim L(k\rho)=(k+1)^{|R^+|}$ by evaluating every formal exponential at one. This evaluation is safe in the final polynomial expression, without dividing by a zero denominator.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 102 / 4 / ii / Solution

{title2=$\operatorname{ch}L(k\rho)$}

For a complex <semisimple Lie algebra> and an <integer> $k\ge0$, the <Weyl character formula> and <Weyl denominator formula> give
$$
\operatorname{ch}L(k\rho)=e^{k\rho}\prod_{\alpha\in R^+}(1+e^{-\alpha}+\cdots+e^{-k\alpha}),\qquad
\dim L(k\rho)=(k+1)^{|R^+|}.
$$
Apply the dilation $e^\mu\mapsto e^{(k+1)\mu}$ to the denominator identity to obtain the numerator. Cancellation leaves a finite <geometric series> for each positive <root of a root system>. Evaluating the resulting polynomial at all formal exponentials equal to one gives the <dimension>, without taking an undefined quotient at the identity.

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 Incoming links: Character of a Weyl-vector multiple