FFT over a fixed and equal numbers
Yes it is true. Your FFT will compute those DFT values for k=0 to k=99.999 all of which are exactly zero but $X[0] = 7000 \cdot 100000 = 7 \times 10^8 $ for k=0. (the DC)  

In order to see why this is the case, we can simply look at the definition of the DFT (as FFT is nothing but a fast algorithm to compute the result of DFT)

The DFT $X[k]$ of a signal $x[n]$ of length N is:

$$ X[k] = \sum_{n=0}^{N-1}{ x[n] e^{-j\frac{2\pi}{N}nk }} ,~~~~~~~~~~~~~~0 \leq n,k \leq N-1 $$

Now assuming the signal $x[n]$ is constant K:
$$x[n] = K ~~~~~~~~~~,~~~~~~ \text{for} ~~~~~~ 0\leq n \leq N-1 $$

The DFT becomes:
$$ X[k] = \sum_{n=0}^{N-1}{ K e^{-j\frac{2\pi}{N}nk }} ,~~~~~~~~~~~~~~0 \leq n,k \leq N-1 $$

There are a number of ways to evaluate or interpret this finite sum of a periodic complex exponential sum all of which will yield the following:
$$X[k]=
\begin{cases}
N \cdot K ~~~~, k = 0 \\
0 ~~~~, 1 \leq k \leq N-1 \\
\end{cases}$$