Why does plain RSA not work with big messages ($m>n$)?
The issue is that we use modular arithmetic. In modular arithmetic, you may view $m \bmod n$ as the remainder of $m$ when divided by $n$. So, for example, $7 \bmod 2 = 1$, also written as $7 \equiv 1 \pmod 2$, because $2(3) + 1 = 7$. 

Now consider what you're guaranteed by $m \equiv c^d \pmod n$, the decryption formula for plain RSA. You're guaranteed that $c^d$ is equivalent to the message modulo $n$. You're not guaranteed that it is precisely equal. Now, if the message is smaller than the modulus &mdash; if $m &lt; n$ &mdash; then by the way we program modular reduction, where we usually give the smallest positive integer modulo $n$ , you're going to get $m$ exactly when you compute $c^d \bmod n$. But if $m&gt;n$, if the message is bigger, you're going to get a smaller integer that is equivalent modulo $n$. 

And indeed this is the case with your example. Note that your example message $m = 33555553$ is larger than your example $n = 4075961$. So when you compute $c^d$, you're going to find an equivalent integer modulo $n$, but not exactly the one you wanted.

In this case, you got back $c^d \bmod n = 947865$. But note what happens if you reduce your original message modulo $n$: you get that your original message is equivalent to $c^d \bmod n$. 

In summary: if your message is larger than your modulus, you won't get the exact message back when you decrypt. You'll get an equivalent message instead. To avoid this, you can instead encrypt a key used in a symmetric cryptosystem (or any cryptosystem that can handle messages of the length you're wanting). As Ricky Demer linked in his comment, this is sometimes called a hybrid cryptosystem. Once you've done this, you just use the other cryptosystem's key to do all of your encryption for you.