so this is why i think an $n$-th order hold is a $\operatorname{rect}\left( \frac{t - T/2}{T} \right)$ convolved against itself $n$ times.
Wikipedia isn't the final reference of all things, but there is something that i sniffed from there. consider sampling and reconstruction (the Shannon Whittaker whatever formula). if the original bandlimited input is $x(t)$ and the samples are $x[n]\triangleq x(nT)$ that bandlimited input can be reconstructed from the samples with
$$ x(t) = \sum\limits_{n=-\infty}^{\infty} x[n] \ \operatorname{sinc}\left( \frac{t - nT}{T} \right) $$
which is the output of an ideal brickwall filter with frequency response:
$$ \begin{align}
H(f) &= \operatorname{rect}(fT) \\
&= \begin{cases}
1 \quad |f| < \frac{1}{2T} \\
0 \quad |f| > \frac{1}{2T} \\
\end{cases}
\end{align}$$
when driven by the ideally sampled function
$$ \begin{align}
x_\text{s}(t) & = x(t) \cdot \sum\limits_{n=-\infty}^{\infty} \delta\left( \frac{t - nT}{T} \right) \\
& = x(t) \cdot T \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\
& = T \sum\limits_{n=-\infty}^{\infty} x(t) \delta(t - nT) \\
& = T \sum\limits_{n=-\infty}^{\infty} x(nT) \delta(t - nT) \\
& = T \sum\limits_{n=-\infty}^{\infty} x[n] \delta(t - nT) \\
\end{align} $$
so when $x_\text{s}(t)$ goes into $H(f)$, what comes out is $x(t)$. the $T$ factor is needed so that the passband gain of the reconstruction filter, $H(f)$ is the dimensionless $1$ or 0 dB.
that means that the impulse response of this ideal brickwall filter is
$$ \begin{align}
h(t) & = \mathcal{F}^{-1} \left\{ H(f) \right\} \\
& = \frac1T \operatorname{sinc}\left( \frac{t}{T} \right) \\
\end{align} $$
the reconstructed $x(t)$ is
$$ x(t) = h(t) \circledast x_\text{s}(t) $$
we clearly cannot realize that reconstruction filter because it is not causal. but with enough delay, we might be able to get closer and closer with a delayed causal $h(t)$.
now a practical DAC does not get particularly close, but because it simply outputs the sample value $x[n]$ for the sample period immediately after the sample, the output of the DAC looks like this
$$ x_\text{DAC}(t) = \sum\limits_{n=-\infty}^{\infty} x[n] \ \operatorname{rect}\left( \frac{t - nT - \tfrac{T}2}{T} \right) $$
and it can be modeled as a filter with impulse response
$$ h_\text{ZOH}(t) = \frac1T \operatorname{rect}\left( \frac{t-\tfrac{T}2}{T} \right) $$
driven by the same $x_\text{s}(t)$. so
$$ x_\text{DAC}(t) = h_\text{ZOH}(t) \circledast x_\text{s}(t) $$
and the frequency response of the implied reconstruction filter is
$$ \begin{align}
H_\text{ZOH}(f) &= \mathcal{F}^{-1}\{ h_\text{ZOH}(t) \} \\
&= \frac{1 - e^{j 2 \pi f T}}{j 2 \pi f T} \\
&= e^{j \pi f T} \operatorname{sinc}(fT) \\
\end{align}$$
note the constant half-sample delay in this frequency response. that's where the Zero-order hold comes from.
so, while the ZOH has the same DC gain as the ideal brickwall reconstruction but not the same gain at other frequencies. in addition, the images in $x_\text{s}(t)$ aren't fully beaten down as would be with the brickwall, but they're beaten down a bit.
so why, in the POV of the time domain, is this? i think it's because of the discontinuities in $x_\text{DAC}(t)$. it's not as bad as the sum of dirac impulses in $x_\text{s}(t)$, but $x_\text{DAC}(t)$ has jump discontinuities.
how do you get rid of jump discontinuities? maybe turn them into discontinuities of the first derivative. and you do that by used if integration in the continuous time domain. so a first-order hold is one where the output of the DAC is run through an integrator with transfer function $\frac{1}{j 2 \pi f T}$ but we try to undo the effects of integrator with a differentiator done in the discrete-time domain. the output of that discrete-time differentiator is $x[n] - x[n-1]$ or Z-transform $X(z) - z^{-1}X(z) = X(z)(1 - z^{-1})$
the transfer function of that differentiator is $(1 - z^{-1})$ or, in the continuous Fourier domain, $(1 - (e^{j 2 \pi f T})^{-1}) = 1 - (e^{-j 2 \pi f T})$. this makes the transfer function of the first-order hold that of the continuous-time integrator, the discrete-time differentiator, and the ZOH of the DAC all multiplied together.
$$ \begin{align}
H_\text{FOH}(f) &= \mathcal{F}^{-1}\{ h_\text{FOH}(t) \} \\
&= \left( \frac{1 - e^{j 2 \pi f T}}{j 2 \pi f T} \right)^2 \\
&= e^{j 2 \pi f T} \operatorname{sinc}^2(fT) \\
\end{align} $$
the impulse response of this is
$$ \begin{align}
h_\text{FOH}(t) &= \mathcal{F}\{ H_\text{FOH}(f) \} \\
&= \left( \operatorname{rect}\left( \frac{t-\tfrac{T}{2}}{T}\right) \right) \circledast \left( \operatorname{rect}\left( \frac{t-\tfrac{T}{2}}{T}\right) \right) \\
&= \frac{1}{T} \operatorname{tri}\left( \frac{t-T}{T} \right) \\
\end{align} $$
now, continuing with this further, the second-order hold would have both continuous zeroth and first derivatives. it does this by integrating again in the continuous-time domain and trying to make up for it in the discrete-time domain with another differentiator. that tosses in another $e^{j \pi f T} \operatorname{sinc}(fT)$ factor which means convolving with another $\operatorname{rect}\left( \frac{t-\tfrac{T}{2}}{T}\right)$.