Probability that rank of a square matrix, whose each element is uniformly random from a ring of integers modulo prime p, is not full I am currently reading a paper from CRYPTO, which is a top conference in cryptology. In the paper, the authors use a thereom like the one given below without a rigorous proof:

the probability that $Rank(A)\neq n$ is upper bounded by $1/p$, where $p$ is a prime, $n$ is an integer, $\mathbb{Z}_p$ is a ring of integers modulo $p$, and $A$ is a square matrix which is chosen from $\mathbb{Z}_p^{n\times n}$ uniformly at random.

I am curious how to rigorously prove the thereom. I would appreciate it if anyone could teach me how.
I'll try to show the probability that the matrix has rank equal to $n$. Two important facts, since $\mathbb{Z}_p$ is a field and not just a ring:


Each column of the matrix can be considered as a random vector drawn from a set of an $n$-dimensional vector space over $\mathbb{Z}_p$, which has $p^n$ vectors.
The number of vectors in a $k$-dimensional subspace will be $p^k$.


So, model the construction of the matrix as drawing $n$ vectors independently at random. For the matrix to have full rank, after the $i$th draw, we need all $i$ vectors to be linearly independent. Assuming the previous $i-1$ vectors are linearly independent, they span a space of dimension $i-1$, so the $i$th vector must be outside of this space. A space of dimension $i-1$ has size $p^{i-1}$, so the probability that a random vector is in that space is $p^{i-1}/p^n$, and we want the opposite to happen, so the probability is:
$$1-\frac{p^{i-1}}{p^n}=1-p^{i-1-n}$$
This even holds for the first vector, which will have rank 1 as long as it is not the all-zero vector.

Thus, the total chance of being full-rank will be 

$$\left(1-p^{-n}\right)\left(1-p^{-n+1}\right)\dots\left(1-p^{-1}\right)$$

This is less than $1-1/p$, which means its converse (that the matrix's rank does not equal $n$) must be at least $1/p$. Are you sure the paper didn't mention a lower bound?

Alternatively, since $(1-p^{-i})\geq 1-1/p$ for $i\geq i$, we can argue that the probability of a full-rank matrix is lower-bounded by 
$$\left(1-\frac{1}{p}\right)^{n} \geq 1-\frac{n}{p}$$

and thus $\frac{n}{p}$ an upper bound on the probability that $Rank(A)\neq n$.