Why is the ring of matrices over a field simple?

Suppose that you have an ideal $\mathfrak{I}$ which contains a matrix with a nonzero entry $a_{ij}$. Multiplying by the matrix that has $0$'s everywhere except a $1$ in entry $(i,i)$, kill all rows except the $i$th row; multiplying by a suitable matrices on the right, kill all columns except the $j$th column; now you have a matrix, necessarily in $\mathfrak{I}$, which contains exactly one nonzero entry, namely $a_{ij}$ in position $(i,j)$.

Now show that $\mathfrak{I}$ must contain all matrices in $M_{n\times n}(k)$. This will show that a $2$-sided ideal consists either of only the $0$ matrix, or must be equal to the entire ring.

Added. Now that you have a matrix that has a single nonzero entry, can you get a matrix that has a single nonzero entry on whatever coordinate you specify, and such that this nonzero entry is whatever element of $k$ you want, by multiplying this matrix (on either left, or right, or both) by suitable elementary matrices? Will they all be in $\mathfrak{I}$?

And...

$$\left(\begin{array}{cc} a&b\\ c&d \end{array}\right) = \left(\begin{array}{cc} a & 0\\ 0 & 0 \end{array}\right) + \cdots$$

A faster, and more general result, which Arturo hinted at, is obtained via following proposition from Grillet's Abstract Algebra, section "Semisimple Rings and Modules", page 360:

Consequence: if $R:=D$ is a division ring, then $M_n(D)$ is simple.

Proof: Suppose there existed an ideal of $M_n(D)$. By the proposition, it'd be of the form $M_n(I)$, for $I\unlhd D$, but division rings do not have any ideals (other than $0$ and $D$), so this is a contradiction. $\blacksquare$