What is the probability that the sum of n numbers between 0 and k is less than k? I was looking at a very recent question, and I wanted to generalize it as 
such:

Suppose that $\{x_1, x_2, \cdots x_n\}$ are n independently uniformly
  distributed real numbers from 0 to k, then what is the probablity that
  their sum is less k?

I want to check if this type of question is solvable using probability and Bayes Theorem as I have attempted below. Thus we require to find:
$$Pr(x_1+x_2+\cdots+x_n\leq k | x_1, x_2, \cdots x_n \in [0, k])$$
$$=\dfrac{Pr(x_1, x_2, \cdots x_n \in [0, k] | x_1+x_2+\cdots+x_n\leq k)Pr(x_1+x_2+\cdots+x_n\leq k)}{Pr(x_1, x_2, \cdots x_n \in [0, k])}$$
Now: 
$$Pr(x_1, x_2, \cdots x_n \in [0, k] | x_1+x_2+\cdots+x_n\leq k) = 1$$
and 
$$Pr(x_1, x_2, \cdots x_n \in [0, k])$$
$$=Pr(x_1\in [0, k])\cdots Pr(x_n\in [0, k]) $$
$$=\dfrac{x_1\cdots x_n}{k^n}$$
The above is true since the $x_i$'s are independent and uniformly distributed between 0 and k.
Now how do you compute: $Pr(x_1+x_2+\cdots+x_n\leq k)$ to find the last term? Is my use of Bayes Theorem correct?
 A: Discrete Case
The number of ways to sum $n$ integers in $[0,k]$ and get $m$ is $\left[x^m\right]\left(x^0+x^1+x^2+\cdots+x^k\right)^n$. Thus, the number of ways to get $m\in[0,k]$ is
$$
\begin{align}
&\sum_{m=0}^k\left[x^m\right]\left(x^0+x^1+x^2+\cdots+x^k\right)^n\\
&=\sum_{m=0}^k\left[x^m\right]\left(\frac{1-x^{k+1}}{1-x}\right)^n\\
&=\sum_{m=0}^k\left[x^m\right]\sum_{j=0}^n(-1)^j\binom{n}{j}x^{j(k+1)}
\sum_{i=0}^\infty(-1)^i\binom{-n}{i}x^i\\
&=\sum_{m=0}^k\left[x^m\right]\sum_{j=0}^n(-1)^j\binom{n}{j}x^{j(k+1)}
\sum_{i=0}^\infty\binom{n+i-1}{i}x^i\\
&=\sum_{m=0}^k\sum_{j=0}^n(-1)^j\binom{n}{j}\binom{n+m-j(k+1)-1}{m-j(k+1)}\\
&=\sum_{m=0}^k\binom{n+m-1}{m}\\
&=\binom{n+k}{k}\\
\end{align}
$$
Therefore, the probability is
$$
\frac1{(k+1)^n}\binom{n+k}{k}
$$

Continuous Case
The continuous case is given by the volume of
$$
\sum_{j=1}^nx_j\le1
$$
where $x_j\ge0$, which is $\frac1{n!}$. The answer is independent of $k$ since the volumes of both the sample space and the success space are multiplied by $k^n$.
Note that the limit of the discrete case as $k\to\infty$ is
$$
\begin{align}
\lim_{k\to\infty}\frac1{(k+1)^n}\binom{n+k}{k}
&=\lim_{k\to\infty}\frac1{(k+1)^n}\frac{(k+n)(k+n-1)\cdots(k+1)}{n!}\\
&=\frac1{n!}
\end{align}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Suppose that $\ds{\braces{x_{1},x_{2},\ldots,x_{n}}}$
  are $\ds{n}$ independently uniformly distributed real numbers from $0$ to $k$, then what is the probablity that their sum is less than $k$ ?.

By definition:
\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{k}{\dd x_{1} \over k}
\int_{0}^{k}{\dd x_{2} \over k}\cdots
\int_{0}^{k}{\dd x_{n} \over k}\bracks{x_{1} + x_{2} + \cdots x_{n} < k}}}
\\[5mm] = &\
{1 \over k^{n}}
\int_{0}^{\infty}\int_{0}^{\infty}\cdots\int_{0}^{\infty}
\bracks{k - x_{1} - x_{2} - \cdots - x_{n} > 0}
\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}\
\pars{\begin{array}{l}
\mbox{Integrals contribution}
\\
\mbox{'beyond'}\ k\ \mbox{vanishes out.}
\end{array}}
\\[5mm] = &\
{1 \over k^{n}}
\int_{0}^{\infty}\int_{0}^{\infty}\cdots\int_{0}^{\infty}
\underbrace{\braces{\int_{c - \infty\ic}^{c + \infty\ic}{\exp\pars{\bracks{k - x_{1} - x_{2} - \cdots - x_{n}}s} \over s}\,{\dd s \over 2\pi\ic}}}
_{\ds{=\ \bracks{k - x_{1} - x_{2} - \cdots - x_{n} > 0}}}\
\dd x_{1}
\,\dd x_{2}\ldots\dd x_{n}
\end{align}

where $\ds{c > 0}$.

Then,
\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{k}{\dd x_{1} \over k}
\int_{0}^{k}{\dd x_{2} \over k}\cdots
\int_{0}^{k}{\dd x_{n} \over k}\bracks{x_{1} + x_{2} + \cdots x_{n} < k}}}
\\ = &\
{1 \over k ^{n}}
\int_{c - \infty\ic}^{c + \infty\ic}{\exp\pars{ks} \over s}\
\underbrace{\pars{\int_{0}^{\infty}\expo{-xs}\dd x}^{n}}
_{\ds{=\ {1 \over s^{n}}}}\ \,{\dd s \over 2\pi\ic} =
{1 \over k ^{n}}\
\overbrace{\int_{c - \infty\ic}^{c + \infty\ic}{\exp\pars{ks} \over s^{n + 1}}
\,{\dd s \over 2\pi\ic}}^{\ds{k^{n} \over n!}}\ =\ \bbx{1 \over n!}
\end{align}
