DFT derivative property?
To get a detailed answer along the lines of what you propose, we need to be careful about the normalization used in discrete Fourier transform (DFT) and inverse discrete Fourier transform (IDFT):
$$\text{DFT: }\quad X[k] = \sum_{n=0}^{N-1} x[n] e^{-j 2\pi kn/N}\tag{1}$$
$$\text{IDFT: }\quad x[n] = \frac{1}{N} \sum_{k=0}^{N-1} X[k] e^{j 2 \pi k n / N}\tag{2}$$
Those have a normalization that is directly compatible with fft and ifft from MATLAB, Octave, NumPy and SciPy. The indexes $k$ and $n$ run from $0$ to $N-1$. (MATLAB and Octave have a different indexing convention, $1$ to $N$.) Then:
$$\begin{array}{c}y[n] = x[n] - x[\operatorname{mod}(n-1, N)]\\
\begin{align}\\
Y &amp;= \operatorname{DFT}\big(y\big)\\
&amp;= \operatorname{DFT}(x*[1, -1, 0, 0, \ldots])\\
&amp;= \operatorname{DFT}(x)\times\operatorname{DFT}([1, -1, 0, 0, \ldots])\\
&amp;= X\times\operatorname{DFT}([1, -1, 0, 0, \ldots])\end{align}\end{array}\tag{3}$$
$$\Rightarrow Y[k]= X[k]\left(1 - e^{-j2\pi k/N}\right),\tag{4}$$
where $\operatorname{mod}$ gives the unsigned remainder, for example $\operatorname{mod}(-1, N) = N - 1$, the symbol $*$ denotes length-$N$ circular convolution and $\times$ denotes multiplication, and all sequences are of length $N$.
Circular convolution in the discrete time domain is exactly equivalent to multiplication in the discrete frequency domain, when DFT and IDFT are used to transform the sequences between the domains. See for example Circular Convolution - MIT OpenCourseWare. For &quot;linear&quot; convolution of discrete sequences, there is no such elegant pair of equivalent operations, which makes me think no expression proposed as an answer to the question will ever be fully satisfactory.
Considering input signals that extend to the left of the output range $0 \le n &lt; N$ of IDFT, calculating the backward difference using modulo indexing is conditionally equal to calculating it without it:
$$x[n] - x[\operatorname{mod}(n-1, N)] = x[n] - x[n-1]\quad\text{conditionally}\tag{5}$$
under the condition that you only calculate it for some of the indexes:
$$0 &lt; n &lt; N,\tag{6}$$
or for $0 \le n &lt; N$ under the condition that there's a hint of periodicity in the signal:
$$x[-1] = x[N-1].\tag{7}$$
A sufficient but not necessary condition that satisfies Eq. 7 is that $x$ is $N$-periodic, which is however prohibited by the signal defined as finite in the question. An example of another condition that satisfies Eq. 7 is $x[-1] = x[N-1] = 0$.
@CedronDawg's first answer provides a formula which corrects in the frequency domain the error in Eq. 5 if neither condition is satisfied: $Y[k] = X[k]\left( 1 - e^{-j2\pi k/N} \right) -  x[-1] + x[N-1]$. As an alternative derivation, in length-$N$ time domain the correction is an impulse:
$$\begin{align}&amp;\big[x[0] - x[−1] - \big(x[0] - x[N-1]\big),\, 0,\, 0,\, \ldots\big]\\
= &amp;\big[x[N-1] - x[-1],\, 0,\, 0,\, \ldots\big],\end{align}\tag{8}$$
which in frequency domain is a constant (see DFT Pairs and Properties: pair row 2, property linearity):
$$x[N-1] - x[-1],\tag{9}$$
to be added to all elements of $Y$ calculated by Eq. 4.
For a general $x$, the condition of Eq. 6 for Eq. 5 enables to use a length $N+1$ DFT and IFT to calculate the backward difference, by shifting the input to the DFT one step to the right, and finally by shifting the output from IDFT one step to the left. With forward difference $x[n + 1] - x[n]$, the shift would not be necessary, and I think this matches your discarding of the 0th sample. For a circular convolution implementation of convolution by a finite sequence, using a longer transform is a common trick to avoid the circular effects in a sufficiently large part of the output of the IDFT. For then obtaining the DFT of a partial IDFT output, in particular a one shorter, I don't think there is any shortcut.
We could perhaps re-express the question as: What is the DFT of the length $N-1$ forward difference of $x$ of length $N$, given $x$ and $X_{N-1} = \operatorname{DFT}(x_{N-1})$, a length $N-1$ DFT of the partial sequence $x_{N-1} = \big[x[0], x[1], \ldots, x[N-2]\big]$? Analogously to Eq. 4 we have:
$$y_{N-1} = x_{N-1}*[-1, 0, 0, \ldots, 0, 0, 1]\tag{10.1}$$
$$\Leftrightarrow Y_{N-1} = X_{N-1}\times\operatorname{DFT}([-1, 0, 0, \ldots, 0, 0, 1])\tag{10.2}$$
$$\Rightarrow Y_{N-1}[k] = X_{N-1}[k]\big(e^{j2\pi k / (N - 1)} - 1\big),\tag{10.3}$$
where each sequence is of length $N-1$. The desired forward difference $f_{N-1}$ is:
$$f_{N-1} = \big[x[1] - x[0],\, x[2] - x[1],\, \ldots,\, x[N-1] - x[N-2]\big].\tag{11}$$
Eq. 10.1 can be expanded to:
$$y_{N-1} = \big[x[1] - x[0],\, x[2] - x[1],\, \ldots,\, x[0] - x[N-2]\big].\tag{12}$$
By comparing Eqs. 11 and 12, it can be seen that:
$$f_{N-1} = y_{N-1} + \big[\ldots,\, 0,\, 0,\, x[N-1] - x[0]\big],\tag{13}$$
where the sequence in brackets is of length $N-1$. Taking the DFT of both sides of Eq. 13 and applying Eq. 10.3 gives the answer:
$$\begin{align}F_{N-1}[k] &amp;= Y_{N-1}[k] + (x[N-1] - x[0])e^{j2\pi k/(N-1)}\\
&amp;=X_{N-1}[k]\big(e^{j2\pi k / (N - 1)} - 1\big) + (x[N-1] - x[0])e^{j2\pi k/(N-1)}.\end{align}\tag{14}$$
This is the length $N-1$ DFT of the length $N-1$ forward difference of $x$ of length $N$.
Alternatively, you might be interested in the derivative of the band-limited signal represented by the samples.