Hardly a research level question, but interesting nonetheless, I hope. Pi is easy, but not e. Where could I start?
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I know a much simpler way; you just need to describe for him/her how to exponentiate. Let $a > b \ge e$ be two numbers (for simplicity, let them be positive integers). The question is, which one is greater: $a^b$, or $b^a$? For instance, let $a=1000$ and $b=999$. Which one is correct: $1000^{999} > 999^{1000}$, or $1000^{999} < 999^{1000}$? (I assume that equality does NOT hold.) It can be shown that $a^b < b^a$. In fact, for every $x \ge 0$, we have $e^x > x^e$ (assuming $x \ne e$). 


Here is how I began my explanation of $e$ to a class of (mostly) freshmen: http://wnk.hamline.edu/~mjhardy/1170/homework/2nd.pdf After #1 and #2, see the words "important punchline". #2 and #3 also have some material relevant to $e$. 


I learned about the "secretary problem" when I was about 10 years old from one of Martin Gardner's books. Though I thought is was cool and amazing, I don't think it gave me much insight into $e$. Here's a way to introduce $e$ with only addition and multiplication, in the form of a game. Tell him he's got a "budget" of say, 100 to work with, and his goal is to pick a bunch of (positive) numbers, not necessarily whole numbers, that add up to 100, where he tries to make the product as large as possible. In his mind, he might first think to break his 100 as 50$\times$50, then realize that 25$\times$25$\times$25$\times$25 is even better, then 10$\times$10$\times$10$\times$10$\times$10$\times$10$\times$10$\times$10$\times$10$\times$10 is even better, and so on. The more numbers you split it into, the better! But wait... 


Does your ten year old understand the slope of a line? If so, try explain to him what a tangent line is by some drawings and examples, and then tell him that the function $e^x$ has the property that the slope of its tangent line at $x$ is $e^x$. I mean, I know this is not a very creative answer, but it does indicate clearly what is so special about $e$. 


When I was this age, I loved the factorials. So why not try to explain it via the Stirling formula? That $n!$ is ``rather close'' to $n^n$ but you need to adjust it a bit via division by $e^n$. Also, if he knows logarithms, the Prime Number Theorem is a good example why $e$ is the ``correct'' base of logarithms. 


If I were to take your title
"How do I explain the number e to a ten year old?" Assume that you are in the (realistic!) situation of having a ten year old in front of you, and you wanting to tell him/her something about $e$. Then it's probably better to try to serve the interests of the ten year old, and not those of the number $e$ (the number $e$ being a meme, it has its own interests). In other words, try to tell him/her something about math that s/he will understand, and will thus appreciate. PS: My guess is that 15 years old is a good age for starting to learn about $e$. It is pointless to try to learn about $e$ before having acquired a deep intuitive understanding of the notion of exponential growth. And I would bet that that's assimilated only relatively late in the cognitive development of a child. 


Thanks for all the answers. Some feedback... Martin, I think the compound interest problem might be a bit beyond him. And J.M., I really don't know how he came across e, but I could ask him next time and report back. Qiaochu and others, I also think the secretary problem is a little bit hard. Todd, I think your method is the best. e simply isn't as "obvious" as Pi, whichever way we choose, but linking it in to spirals is by far the most engaging. I'll give it a go and let you know how I get all. Thanks all of you for your answers. 


If you can explain $\pi$, then you can also explain radians, and this leads to a geometric interpretation of $e$ that is not in the wikipedia article Martin Brandenburg referred to. Get a big sheet of paper and make a polar coordinate grid, and ask your tenyearold to put a pencil down somewhere away from the origin, and try to draw a continuous curve that meets each radial line in a 45 degree angle. The curve will be a logarithmic spiral that spirals in toward the origin. Next, have the child trace along the curve toward the origin, starting at any point and stopping after 1 radian's worth of angular measure has been traversed. How much further away from the origin is the starting point than the stopping point? The answer is: by a factor of $e$. (By similarity considerations, it doesn't matter where you start.) You could also make a little story out of this by getting the child to imagine four ants placed on the corners of a square, each facing its neighbor in the clockwise direction when looking down at them, and then each chasing its neighbor all moving at the same speed. The trajectories they trace out are logarithmic spirals which tend toward the origin, and which cut radial lines at 45 degree angles. I'll leave remaining details of the story to you. In conjunction with such experiments, you could examine spirals on seashells (which are approximately logarithmic spirals), and notice that similarity of the shape of the spiral no matter how you turn it. Again, the spirals cut a "radial line" invariably at the same angle. You can then perform thought experiments on the sea shell along the same lines as above. 


Via percents in a bank? If we have a bank deposit for $x$ per annum, then we get from $N$ dollars $N(1+x)$ after a year. But if we get $x/12$ every month, we get $N(1+x/12)^{12}$ after a year. If we get $x/365$ every day, we get $N(1+x/365)^{365}$ after year, and if we get our percents continuously, we get $Ne^x$ after a year. 


What about the basic (1+1/10)^10, (1+1/100)^100, and so on ? 


There is a nice way that I learned from Martin Gardner's books when I was young(er). Imagine the following situation. There is a party with $N$ people. All of them throw their hats in the middle of the room, and then each of them takes one hat randomly. What is the probability that nobody gets its own hat? The surprising answer is that this probability equals $11+\frac{1}{2!}\frac{1}{3!} + \cdots + \frac{(1)^N}{N!}$, which goes to $\frac{1}{e}=0.36788...$ for $N \to \infty$. 


Here is one way which I learned from Clio Cresswell's Mathematics and Sex, although unfortunately I'm not sure how to prove it. Suppose you are sure that you will meet exactly $n$ suitable marriage partners in your life, but you don't know which one is the right one for you. As you meet each of your marriage partners in turn, you can choose to marry or reject them, and you cannot return to a marriage partner you have previously rejected. All you can do is maintain a list of how suitable each past marriage partner has been and compare that to your current suitor. What is the optimum strategy to maximize your marital bliss? As it turns out, the answer (under these somewhat artificial conditions) is to reject the first $\frac{n}{e}$ of your partners, then marry the next partner after that which is as suitable or more suitable than your previous partners. 

