Two optimizations speed up the process:
First, when you compute $m^e \bmod n$ you don't do it by computing $m^e$ then reducing it $\bmod n$. You do it by reducing $\bmod n$ after each multiplication in the computation of $m^e$.
Second, you don't work your way up from $m$ to $m^e$ by stepping through all the powers of $m$. You compute $m^2$ with one multiplication, then $m^4 = m^2 m^2$ with another multiplication, then $m^8 = m^4 m^4$, and so on, reaching $m^{2^k}$ after $k$ steps. Using the binary representation of $e$, you select a subset of $\{m, m^2, ..., m^{2^k}\}$ whose product is $m^e$. This is called the "square and multiply" method.
Every intermediate result in the computation is immediately reduced modulo $n$, and the only operation you perform is multiplication, so you never see a number bigger than $n^2$, which is way smaller than the unreduced $m^e$ when the numbers are of the size needed for practical cryptography.
Some programming languages contain a "modular exponentiation" function which takes the 3 arguments $m$, $e$, and $n$ and returns $m^e \bmod n$ using the above method. When working with a lower-level language that doesn't include it, you will write it first. (It's not hard if you already have the big integer multiplication taken care of.) Since I see this question originated on the mathematica site, here's the function in mathematica: PowerMod
I don't see a modexp button on your javascript calculator, so it's really not the right tool to use here (unless you want to work your way through the square and multiply method by hand - maybe that's good to do once to get the idea burned into your brain, but after that it'll be boring)
I've looked at the two papers. Clearly, strength is not the point, even $p,q$ of the order of about a hundred are open to attacks by hand/calculator.
Firstly $e\neq d$ is an absolute requirement for the definition of RSA since in practice $e$ is public and $d$ is private (needs to be kept secret) so both cannot be the same.
The other problem is that one chooses odd primes for RSA. The reason being if you choose $p=2,$ then your $\phi(n)=(p-1)(q-1)=q-1,$ directly reveals the value of $p$. The reason this is important is that the valid user in posession of the secret key $e,$ also knows $p$ and $q$ (it's been proved that knowing $(n,e)$ you can find $p,q$) and thus can break the system.
In fact the exponentiation for decryption $C^d$ are performed separately mod $p$ and $q$ by the legitimate user for efficiency, and then the Chinese Remainder Theorem is used to obtain $C^d$ modulo $n.$
What if we chose the smaller prime to be $p=3$? This presents a problem for the popular choice of exponent $e=3.$ By Fermat's little theorem, $$a^p\equiv a~(\textrm{mod}~p),$$ for all integers $a$ which means that the subgroup of multiplication modulo $p=3,$ is "inactive" in some sense in the exponentiation operation and then $q$ could be discovered by a cryptanalyst.
In summary, the smaller prime must be at least $p=5.$ Taking $p=5,$ the smallest $q$ would be 7, otherwise taking the integer square root of $n$ is equivalent to factorisation.
If $p=5,q=7$ then $\phi(n)=24,$ which violates the requirement $gcd(e,\phi(n))=1,$ for the exponent $e=5.$
So take $p=5,q=11$ to get $n=55,$ ruling out $n=33,$ as I explained above.
Best Answer
Let we have RSA encrypt $E(x) = x^{49} \bmod 215629$ then the double RSA encrypt is \begin{align} E^2(x) &= (x^{49})^{49} \bmod 215629\\ &= \color{red}{x^{49\cdot49}} \bmod 215629\\ &= x^{49^2} \bmod 215629\\ &= x^{2401} \bmod 215629\\ \end{align}
Now, if you encrypt 10 times than
\begin{align} E^{10}(x) &= ((x^{49})^{{49}^{\cdots^{49}}} ) \bmod 215629\\ &= x^{49^{10}} \bmod 215629\\ &= x^{79792266297612001} \bmod 215629\\ \end{align}
From Euler's theorem we know that $a^{\varphi(n)} \equiv 1 \pmod n$
The $\varphi(383\cdot 563) = (383-1)\cdot (563-1)= 214684$
With a simple test $$79792266297612001 = 1 \bmod 214684$$