= Solution
Taking $k_2\to k_1=k$ in the preceding <probability density> gives $p_2(t)=k^2te^{-kt}$. Convolving once more with an independent <exponential distribution> gives
$$
p_3(t)=\int_0^t k^2s e^{-ks}ke^{-k(t-s)}\,ds=\frac{k^3t^2}{2}e^{-kt}.
$$
More generally, suppose $p_n(s)=k^ns^{n-1}e^{-ks}/(n-1)!$. Then
$$
p_{n+1}(t)=\frac{k^{n+1}e^{-kt}}{(n-1)!}\int_0^t s^{n-1}\,ds
=\frac{k^{n+1}t^n e^{-kt}}{n!}.
$$
Since $p_1(t)=ke^{-kt}$, this proves the <Erlang distribution> formula by <mathematical induction>. Normalization follows either from the <convolution> or $\int_0^\infty t^{n-1}e^{-kt}\,dt=(n-1)!/k^n$.
$$
\boxed{p_n(t)=\frac{k^nt^{n-1}e^{-kt}}{(n-1)!},\qquad
\Pr(t<T<t+dt)=p_n(t)dt+o(dt),\qquad
t_{\rm mode}=\frac{n-1}{k}.}
$$
For $n>1$, differentiating $\log p_n$ gives $(n-1)/t-k=0$, and the derivative changes from positive to negative. For $n=1$ the maximum occurs at the boundary $t=0$. \b[Source dimensional qualification:] the printed expression without $dt$ is a <probability density>, with units of inverse time; it cannot itself be a dimensionless interval <probability>.
In the right panel of the original illustration in <2/d>, $p_2/k$ rises linearly near zero and peaks at $kt=1$ with height $1/e$. The curve $p_3/k$ rises quadratically and peaks later, at $kt=2$, with lower height $2e^{-2}$. Both have exponential tails, but the latter has the larger <expected value> $3/k$ and <variance> $3/k^2$, compared with $2/k$ and $2/k^2$. This distinguishes both the onset and the width of the requested sketches.
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