Solving a factorial equation Prove that the only solution to $k! +m! =n!$ is
$k = 1, m = 1, n = 2$
How would you go about this? I can't seem to figure out where to start.

If you assume, wolog, $k\le m$ and you divide by $k!$ you get:
$1 +  [(k+1).....m]= [(k+1).....m][(m+1)....n]$
Which is only possilbe if $[(k+1).....m] =1$.
Which is $k < m$ means $k=0$ and $m=1$.
......
Just finesse a bit allow for $k = m$ and divide allow "$[(k+1).....m]$" to be $1$ you get, if you divide by $k!=m!$:
$1 + 1= [(m+1)....n] = 2$
which means either $n = m+1=2$ or $2=(m+1)(m+2)=(m+1)n$
which means $m=k=1$ and $n=2$ or $m=k=0$ and $n=2$. 
So only solutions are $0! + 0! = 2=2!$ 
$0! + 1! = 2!$
or 
$1! + 1! = 2!$.
