Brocard's problem is the solution of the equation, 𝒏!+𝟏= 𝒎𝟐, where m and n are natural numbers. So far only 3 solutions have been found, namely (n,m) = (4,5), (5,11), and (7,71). The purpose of this paper is to show that there are no other solutions. Firstly, it will be shown that if (n,m) is to be a solution to Brocard's problem, then n! = 4AB, where A is even, B is odd, and |A – B| = 1. If n is even (n = 2x) and > 4, it will be shown that necessarily 𝑨=(𝟐𝒙)‼𝟒𝒚 and 𝑩=𝒚(𝟐𝒙−𝟏)‼, for some odd y > 1. Next, it will be shown that x < 2y, and this leads to an inequality in x [namely, (𝒙(𝟐𝒙−𝟏)‼ ± 𝟏)𝟐−𝟏−(𝟐𝒙)!<𝟎], for which there is no solution when x ≥ 3. If n is odd, there is a similar procedure.